算法设计与分析讨论班上机作业

2018年4月2日8,5320

凸包

A:Wall POJ1113

求凸包周长加一个圆

#include<map>
#include<deque>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define inf 2000000000
using namespace std;
int n,R,top;
struct P{
    double x,y;
}p[1005],s[1005];
double dis(P a,P b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
P operator-(P a,P b)
{
    P t;t.x=a.x-b.x;t.y=a.y-b.y;return t;
}
double operator*(P a,P b)
{
    return a.x*b.y-a.y*b.x;
}
bool operator<(P a,P b)
{
    double t=(a-p[1])*(b-p[1]);
    if(t==0)return dis(a,p[1])<dis(b,p[1]);
    return t<0;
}
void graham()
{
    int k=1;
    for(int i=2;i<=n;i++)
        if(p[k].y>p[i].y||(p[k].y==p[i].y&&p[k].x>p[i].x))
            k=i;
    swap(p[1],p[k]);
    sort(p+2,p+n+1);
    s[++top]=p[1];s[++top]=p[2];
    for(int i=3;i<=n;i++)
    {
        while(top>1&&(p[i]-s[top-1])*(s[top]-s[top-1])<=0)
            top--;
        s[++top]=p[i];
    }
}
int main()
{
    scanf("%d%d",&n,&R);
    for(int i=1;i<=n;i++)
        scanf("%lf%lf",&p[i].x,&p[i].y);
    graham();
    double ans=0;
    s[0]=s[top];
    for(int i=1;i<=top;i++)
        ans+=sqrt(dis(s[i],s[i-1]));
    ans=ans+2*acos(-1)*R;
    printf("%d\n",(int)round(ans));
    return 0;
}

#include<map>

#include<deque>

#include<ctime>

#include<cmath>

#include<cstdio>

#include<vector>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define inf 2000000000

using namespace std;

int n,R,top;

struct P{

double x,y;

}p[1005],s[1005];

double dis(P a,P b)

{

return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);

}

P operator-(P a,P b)

{

P t;t.x=a.x-b.x;t.y=a.y-b.y;return t;

}

double operator*(P a,P b)

{

return a.x*b.y-a.y*b.x;

}

bool operator<(P a,P b)

{

double t=(a-p[1])*(b-p[1]);

if(t==0)return dis(a,p[1])<dis(b,p[1]);

return t<0;

}

void graham()

{

int k=1;

for(int i=2;i<=n;i++)

if(p[k].y>p[i].y||(p[k].y==p[i].y&&p[k].x>p[i].x))

k=i;

swap(p[1],p[k]);

sort(p+2,p+n+1);

s[++top]=p[1];s[++top]=p[2];

for(int i=3;i<=n;i++)

{

while(top>1&&(p[i]-s[top-1])*(s[top]-s[top-1])<=0)

top--;

s[++top]=p[i];

}

int main()

{

scanf("%d%d",&n,&R);

for(int i=1;i<=n;i++)

scanf("%lf%lf",&p[i].x,&p[i].y);

graham();

double ans=0;

s[0]=s[top];

for(int i=1;i<=top;i++)

ans+=sqrt(dis(s[i],s[i-1]));

ans=ans+2*acos(-1)*R;

printf("%d\n",(int)round(ans));

return 0;

}

B:Scrambled Polygon POJ2007

排序凸包上的点

#include<set>
#include<map>
#include<deque>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define inf 2000000000
using namespace std;
int n;
struct P{
    int x,y;
}p[10005];
int dis(P a,P b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
P operator-(P a,P b)
{
    P t;t.x=a.x-b.x;t.y=a.y-b.y;return t;
}
int operator*(P a,P b)
{
    return a.x*b.y-a.y*b.x;
}
bool operator<(P a,P b)
{
    int t=(a-p[1])*(b-p[1]);
    return t>0;
}
int main()
{
    n=1;
    while(scanf("%d%d",&p[n].x,&p[n].y)!=EOF)
        n++;
    n--;
    sort(p+2,p+n+1);
    for(int i=1;i<=n;i++)
        printf("(%d,%d)\n",p[i].x,p[i].y);
    return 0;
}

#include<set>

#include<map>

#include<deque>

#include<ctime>

#include<cmath>

#include<cstdio>

#include<vector>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define inf 2000000000

using namespace std;

int n;

struct P{

int x,y;

}p[10005];

int dis(P a,P b)

{

return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);

}

P operator-(P a,P b)

{

P t;t.x=a.x-b.x;t.y=a.y-b.y;return t;

}

int operator*(P a,P b)

{

return a.x*b.y-a.y*b.x;

}

bool operator<(P a,P b)

{

int t=(a-p[1])*(b-p[1]);

return t>0;

}

int main()

{

n=1;

while(scanf("%d%d",&p[n].x,&p[n].y)!=EOF)

n++;

n--;

sort(p+2,p+n+1);

for(int i=1;i<=n;i++)

printf("(%d,%d)\n",p[i].x,p[i].y);

return 0;

}

动态规划 G 题真的坑

A:Fourier’s Lines POJ1923

#include<set>
#include<map>
#include<deque>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define inf 10000000000
#define ll long long
using namespace std;
int n, m;
int f[105][10005];
int check(int n, int m)
{
        if(m && (m < n - 1 || m > n * (n - 1) / 2))
                return 0;
        if(m == 0)return 1;
        if(f[n][m] != -1)return f[n][m];        
        f[n][m] = 0;
        for(int i = 1; i <= n; i++)
                f[n][m] |= check(n - i, m - i * (n - i));
        return f[n][m];
}
int main()
{
        memset(f, -1, sizeof(f));
        int T = 0;
        while(scanf("%d%d", &n, &m))
        {
                T++;
                if(n == 0)break;
                if(check(n, m))
                        printf("Case %d: %d lines with exactly %d crossings can cut the plane into %d pieces at most.\n", T, n, m, n + m + 1);
                else
                        printf("Case %d: %d lines cannot make exactly %d crossings.\n", T, n, m);                        
        }                
        return 0;
}

#include<set>

#include<map>

#include<deque>

#include<ctime>

#include<cmath>

#include<cstdio>

#include<vector>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define inf 10000000000

#define ll long long

using namespace std;

int n, m;

int f[105][10005];

int check(int n, int m)

{

if(m && (m < n - 1 || m > n * (n - 1) / 2))

return 0;

if(m == 0)return 1;

if(f[n][m] != -1)return f[n][m];

f[n][m] = 0;

for(int i = 1; i <= n; i++)

f[n][m] |= check(n - i, m - i * (n - i));

return f[n][m];

}

int main()

{

memset(f, -1, sizeof(f));

int T = 0;

while(scanf("%d%d", &n, &m))

{

T++;

if(n == 0)break;

if(check(n, m))

printf("Case %d: %d lines with exactly %d crossings can cut the plane into %d pieces at most.\n", T, n, m, n + m + 1);

else

printf("Case %d: %d lines cannot make exactly %d crossings.\n", T, n, m);

}

return 0;

}

B:Tour POJ2677

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#define inf 1000000000
using namespace std;
int n;
double f[55][55];
struct data{
    int x, y;
}p[55];
double dis(data a, data b)
{
    return sqrt((a.x - b.x)*(a.x - b.x)+(a.y - b.y)*(a.y - b.y));
}
int main()
{
    while(cin >> n)
    {
        for(int i = 1; i <= n; i++)
            cin >> p[i].x >> p[i].y;
        memset(f, 127, sizeof(f));
        f[1][1] = 0;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j < i; j++)
            {
                f[i][j] = min(f[i][j], f[i - 1][j] + dis(p[i - 1], p[i]));
                f[i][i - 1] = min(f[i][i - 1], f[i - 1][j] + dis(p[j], p[i]));
            }
        printf("%.2lf\n", f[n][n-1] + dis(p[n-1], p[n]));
    }
    return 0;
}

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#define inf 1000000000

using namespace std;

int n;

double f[55][55];

struct data{

int x, y;

}p[55];

double dis(data a, data b)

{

return sqrt((a.x - b.x)*(a.x - b.x)+(a.y - b.y)*(a.y - b.y));

}

int main()

{

while(cin >> n)

{

for(int i = 1; i <= n; i++)

cin >> p[i].x >> p[i].y;

memset(f, 127, sizeof(f));

f[1][1] = 0;

for(int i = 1; i <= n; i++)

for(int j = 1; j < i; j++)

{

f[i][j] = min(f[i][j], f[i - 1][j] + dis(p[i - 1], p[i]));

f[i][i - 1] = min(f[i][i - 1], f[i - 1][j] + dis(p[j], p[i]));

}

printf("%.2lf\n", f[n][n-1] + dis(p[n-1], p[n]));

}

return 0;

}

C:Increasing Sequences POJ1239

#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#define inf 1000000000
using namespace std;
int n;
char s[105];
int f[105][105]; // 某个数是[l, r], 其能获得的最小的末尾数是[f[l][r], n]
vector<string> ans;
bool cmp(int l1, int r1, int l2, int r2) // 1 > 2?
{
   int p1 = l1, p2 = l2;
    while(s[p1] == '0' && p1 <= r1)p1++;
    while(s[p2] == '0' && p2 <= r2)p2++;
    if(r1 - p1 - 1 != r2 - p2 - 1)
        return (r1 - p1) > (r2 - p2);
    if(r1 - p1 + 1 == 0)return 0;
    while(p1 <= r1){
        if(s[p1] > s[p2])return 1;
        if(s[p1] < s[p2])return 0;
        p1++;
        p2++;
    }
    return 0;
}
int dp(int l, int r)
{
    if(r == n)f[l][r] = l;
    if(f[l][r])return f[l][r];
    f[l][r] = l;
    for(int i = r + 1; i <= n; i++)
        if(dp(r + 1, i) && cmp(r + 1, i, l, r))            
            f[l][r] = max(f[l][r], f[r + 1][i]);
    return f[l][r];
}
 
void print(int last, int x)
{
    if(x == n)return;
    string dig = "";
    int pos = n;
    for(int i = x + 1; i <= n; i++)
        if((!last || cmp(x + 1, i, last, x))
           &&
           (!cmp(f[x + 1][i], n, f[x + 1][pos], n)))
            pos = i;
           // 下一段是 [x + 1, pos]
    for(int i = x + 1; i <= pos; i++)
        dig = dig + s[i];
    ans.push_back(dig);
    print(x + 1, pos);
}
int main()
{    
    while(scanf("%s", s + 1))
    {
        n = strlen(s + 1);
        if(n == 1 && s[1] == '0')break;
        memset(f, 0, sizeof(f));
        for(int i = 1; i <= n; i++)
            for(int j = i; j <= n; j++)
                dp(i, j);
        ans.clear();
        print(1, 0);
        cout << ans[0];
        for(int i = 1; i < ans.size(); i++)
            cout << "," << ans[i];
        printf("\n");
    }
    return 0;
}

#include <cmath>

#include <cstdio>

#include <vector>

#include <cstring>

#include <iostream>

#define inf 1000000000

using namespace std;

int n;

char s[105];

int f[105][105]; // 某个数是[l, r], 其能获得的最小的末尾数是[f[l][r], n]

vector<string> ans;

bool cmp(int l1, int r1, int l2, int r2) // 1 > 2?

{

int p1 = l1, p2 = l2;

while(s[p1] == '0' && p1 <= r1)p1++;

while(s[p2] == '0' && p2 <= r2)p2++;

if(r1 - p1 - 1 != r2 - p2 - 1)

return (r1 - p1) > (r2 - p2);

if(r1 - p1 + 1 == 0)return 0;

while(p1 <= r1){

if(s[p1] > s[p2])return 1;

if(s[p1] < s[p2])return 0;

p1++;

p2++;

}

return 0;

}

int dp(int l, int r)

{

if(r == n)f[l][r] = l;

if(f[l][r])return f[l][r];

f[l][r] = l;

for(int i = r + 1; i <= n; i++)

if(dp(r + 1, i) && cmp(r + 1, i, l, r))

f[l][r] = max(f[l][r], f[r + 1][i]);

return f[l][r];

}

void print(int last, int x)

{

if(x == n)return;

string dig = "";

int pos = n;

for(int i = x + 1; i <= n; i++)

if((!last || cmp(x + 1, i, last, x))

(!cmp(f[x + 1][i], n, f[x + 1][pos], n)))

pos = i;

// 下一段是 [x + 1, pos]

for(int i = x + 1; i <= pos; i++)

dig = dig + s[i];

ans.push_back(dig);

print(x + 1, pos);

}

int main()

{

while(scanf("%s", s + 1))

{

n = strlen(s + 1);

if(n == 1 && s[1] == '0')break;

memset(f, 0, sizeof(f));

for(int i = 1; i <= n; i++)

for(int j = i; j <= n; j++)

dp(i, j);

ans.clear();

print(1, 0);

cout << ans[0];

for(int i = 1; i < ans.size(); i++)

cout << "," << ans[i];

printf("\n");

}

return 0;

}

D:Charlie’s Change POJ1787

#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#define inf 1000000000
using namespace std;
int P;
int f[5][10005], g[5][10005];
int w[5] = {0, 1, 5, 10, 25}, num[5];
int ans[5];
int main()
{
        while(scanf("%d", &P))
        {
                if(P == 0)return 0;
                memset(f, -1, sizeof(f));
                memset(g, 0, sizeof(g));
                for(int i = 1; i <= 4; i++)
                        scanf("%d", &num[i]);
                f[0][0] = 1;
                for(int i = 1; i <= 4; i++)
                        for(int j = 0; j <= P; j++)
                        {
                                f[i][j] = f[i - 1][j];
                                if(j >= w[i] && f[i][j - w[i]] != -1 && f[i][j - w[i]] + 1 > f[i][j] && g[i][j - w[i]] + 1 <= num[i])
                                {
                                        f[i][j] = f[i][j - w[i]] + 1;
                                        g[i][j] = g[i][j - w[i]] + 1;
                                }
                        }
                if(f[4][P] == -1)
                {
                        puts("Charlie cannot buy coffee.");
                        continue;
                }
                for(int i = 4; i >= 1; i--)
                {
                        ans[i] = g[i][P];
                        P -= ans[i] * w[i];
                }
                printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n", ans[1], ans[2], ans[3], ans[4]);
        }
        return 0;
}

#include <cmath>

#include <cstdio>

#include <vector>

#include <cstring>

#include <iostream>

#define inf 1000000000

using namespace std;

int P;

int f[5][10005], g[5][10005];

int w[5] = {0, 1, 5, 10, 25}, num[5];

int ans[5];

int main()

{

while(scanf("%d", &P))

{

if(P == 0)return 0;

memset(f, -1, sizeof(f));

memset(g, 0, sizeof(g));

for(int i = 1; i <= 4; i++)

scanf("%d", &num[i]);

f[0][0] = 1;

for(int i = 1; i <= 4; i++)

for(int j = 0; j <= P; j++)

{

f[i][j] = f[i - 1][j];

if(j >= w[i] && f[i][j - w[i]] != -1 && f[i][j - w[i]] + 1 > f[i][j] && g[i][j - w[i]] + 1 <= num[i])

{

f[i][j] = f[i][j - w[i]] + 1;

g[i][j] = g[i][j - w[i]] + 1;

}

if(f[4][P] == -1)

{

puts("Charlie cannot buy coffee.");

continue;

}

for(int i = 4; i >= 1; i--)

{

ans[i] = g[i][P];

P -= ans[i] * w[i];

}

printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n", ans[1], ans[2], ans[3], ans[4]);

}

return 0;

}

E:Strange Towers of Hanoi POJ1958

#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#define inf 1000000000
using namespace std;
int f[15];
int main()
{
        f[1] = 1;
        for(int i = 2; i <= 12; i++)
        {
                f[i] = 10000;
                for(int k = 1; k <= i; k++)
                        f[i] = min(f[i], 2 * f[i - k] + (int)pow(2, k) - 1);
        }
        for(int i = 1; i <= 12; i++)
                printf("%d\n", f[i]);
        return 0;
}

#include <cmath>

#include <cstdio>

#include <vector>

#include <cstring>

#include <iostream>

#define inf 1000000000

using namespace std;

int f[15];

int main()

{

f[1] = 1;

for(int i = 2; i <= 12; i++)

{

f[i] = 10000;

for(int k = 1; k <= i; k++)

f[i] = min(f[i], 2 * f[i - k] + (int)pow(2, k) - 1);

}

for(int i = 1; i <= 12; i++)

printf("%d\n", f[i]);

return 0;

}

F:Balanced Lineup POJ3274

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n,q,x,y,a[50001],mn[50001][16],mx[50001][16];
void pre()
{
    for(int i=1;i<=n;i++)
        mx[i][0]=mn[i][0]=a[i];
    int t=log(n)/log(2);
    for(int i=1;i<=t;i++)
        for(int j=n;j>0;j--)
        {
            mx[j][i]=mx[j][i-1];
            if(j+(1<<(i-1))<=n)
                mx[j][i]=max(mx[j][i],mx[j+(1<<(i-1))][i-1]);
        }
    for(int i=1;i<=t;i++)
        for(int j=n;j>0;j--)
        {
            mn[j][i]=mn[j][i-1];
            if(j+(1<<(i-1))<=n)
                mn[j][i]=min(mn[j][i],mn[j+(1<<(i-1))][i-1]);
        }
}
int rmq(int l,int r)
{
    int m=log(r-l+1)/log(2);
    int a=max(mx[l][m],mx[r-(1<<m)+1][m]);
    int b=min(mn[l][m],mn[r-(1<<m)+1][m]);
    return a-b;
}
int main()
{
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    pre();
    while(q--)
    {
        scanf("%d%d",&x,&y);
        printf("%d\n",rmq(x,y));
    }
    return 0;
}

#include<iostream>

#include<cstdio>

#include<cmath>

using namespace std;

int n,q,x,y,a[50001],mn[50001][16],mx[50001][16];

void pre()

{

for(int i=1;i<=n;i++)

mx[i][0]=mn[i][0]=a[i];

int t=log(n)/log(2);

for(int i=1;i<=t;i++)

for(int j=n;j>0;j--)

{

mx[j][i]=mx[j][i-1];

if(j+(1<<(i-1))<=n)

mx[j][i]=max(mx[j][i],mx[j+(1<<(i-1))][i-1]);

}

for(int i=1;i<=t;i++)

for(int j=n;j>0;j--)

{

mn[j][i]=mn[j][i-1];

if(j+(1<<(i-1))<=n)

mn[j][i]=min(mn[j][i],mn[j+(1<<(i-1))][i-1]);

}

int rmq(int l,int r)

{

int m=log(r-l+1)/log(2);

int a=max(mx[l][m],mx[r-(1<<m)+1][m]);

int b=min(mn[l][m],mn[r-(1<<m)+1][m]);

return a-b;

}

int main()

{

scanf("%d%d",&n,&q);

for(int i=1;i<=n;i++)

scanf("%d",&a[i]);

pre();

while(q--)

{

scanf("%d%d",&x,&y);

printf("%d\n",rmq(x,y));

}

return 0;

}

G:Banal Tickets POJ1608

#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#define inf 1000000000
#define rad 100000000
#define ll long long
using namespace std;
int n;
char str[100];
ll f[2][55][40][20][20], g[55][40][20][20]; // i 2 3 5 7 0?
  
struct data{
    int l;ll v[20];
    data(){
        l=1;
        memset(v,0,sizeof(v));
    }
    data(ll x){
        memset(v,0,sizeof(v));
        l=0;
        while(x)
        {
            v[++l]=x%rad;
            x/=rad;
        }
    }
    ll& operator[](int x){
        return v[x];
    }
    friend data operator+(data a,data b){
        if(a.l<b.l)swap(a,b);
        for(int i=1;i<=b.l;i++)
            a[i]+=b[i];
        for(int i=1;i<=a.l;i++)
            if(a[i]>=rad)
            {
                if(i==a.l)a.l++;
                a[i+1]+=a[i]/rad;
                a[i]%=rad;
            }
        return a;
    }
    friend data operator-(data a,data b){
        for(int i=1;i<=b.l;i++)
            a[i]-=b[i];
        for(int i=1;i<=a.l;i++)
            if(a[i]<0)
            {
                a[i]+=rad;
                a[i+1]--;
            }
        while(!a[a.l]&&a.l>1)a.l--;
        return a;
    }
    friend data operator*(data a,ll b){
        for(int i=1;i<=a.l;i++)
            a[i]*=b;
        for(int i=1;i<=a.l;i++)
            if(a[i]>=rad)
            {
                a[i+1]+=a[i]/rad;
                a[i]%=rad;
                if(i==a.l)a.l++;
            }
        return a;
    }
    friend data operator*(data a,data b){
        data c;
        for(int i=1;i<=a.l;i++)
            for(int j=1;j<=b.l;j++)
                c[i+j-1]+=a[i]*b[j];
        for(int i=1;i<=a.l+b.l;i++)
        {
            if(c[i]>=rad)
            {
                c[i+1]+=c[i]/rad;
                c[i]%=rad;
            }
            if(c.v[i])c.l=i;
        }
        return c;
    }
    friend void print(data a){
        printf("%lld",a[a.l]);
        for(int i=a.l-1;i;i--)printf("%.8lld",a[i]);
        puts("");
    }
};
  
void dp(int L, int R)
{
    memset(f, 0, sizeof(f));        
    int p = 0, q = 1;
    f[0][0][0][0][0] = 1;
    for(int x = L - 1; x <= R - 1; x++)
    {
        int l = x - L + 1;
        char ch = str[x + 1];
        for(int a = 0; a <= 3 * l; a++)
            for(int b = 0; b <= 2 * l; b++)
                for(int c = 0; c <= l; c++)
                    for(int d = 0; d <= l; d++)
                        if(f[p][a][b][c][d])
                        {
                            if(ch == '1' || ch == '?')
                                f[q][a][b][c][d] += f[p][a][b][c][d];
                            if(ch == '2' || ch == '?')
                                f[q][a+1][b][c][d] += f[p][a][b][c][d];
                            if(ch == '3' || ch == '?')
                                f[q][a][b+1][c][d] += f[p][a][b][c][d];
                            if(ch == '4' || ch == '?')
                                f[q][a+2][b][c][d] += f[p][a][b][c][d];
                            if(ch == '5' || ch == '?')
                                f[q][a][b][c+1][d] += f[p][a][b][c][d];
                            if(ch == '6' || ch == '?')
                                f[q][a+1][b+1][c][d] += f[p][a][b][c][d];
                            if(ch == '7' || ch == '?')
                                f[q][a][b][c][d+1] += f[p][a][b][c][d];
                            if(ch == '8' || ch == '?')
                                f[q][a+3][b][c][d] += f[p][a][b][c][d];
                            if(ch == '9' || ch == '?')
                                f[q][a][b+2][c][d] += f[p][a][b][c][d];
                        }                
        memset(f[p], 0, sizeof(f[p]));
        swap(p, q);
    }
}
data qpow(ll t, ll b)
{
    data ans = data(1);
    data a = (data)(t);
    for(ll i = b; i; i >>= 1, a = a * a)
        if(i & 1)ans = ans * a;
    return ans;
}
data calzero()
{
    int ans1 = 0, ans2 = 0;
    for(int i = 1; i <= n; i++)
        if(str[i] == '?')ans1++;
    for(int i = n + 1; i <= 2 * n; i++)
        if(str[i] == '?')ans2++;
    data t1 = qpow(10, ans1) - qpow(9, ans1);
    data t2 = qpow(10, ans2) - qpow(9, ans2);
    return t1 * t2;
}
int main()
{
    scanf("%d", &n);
    scanf("%s", str+1);
    ll t1 = 0, t2 = 0, zl = 0, zr = 0;
    for(int i = 1; i <= 2 * n; i++)
        if(str[i] == '?')
        {
            if(i <= n)t1++;
            else t2++;
        }
        else if(str[i] == '0')
        {
            if(i <= n)zl = 1;
            else zr = 1;
        }
    data ans;
    if(zl || zr)
    {
        if(zl && zr)
            ans = qpow(10, t1 + t2);
        else if(zl)
            ans = qpow(10, t1) * (qpow(10, t2) - qpow(9, t2));
        else ans = (qpow(10, t1) - qpow(9, t1)) * qpow(10, t2);
    }
    else
    {
        dp(1, n);
        memcpy(g, f[n&1], sizeof(f[n&1]));
        dp(n + 1, 2 * n);
        for(int a = 0; a <= 3 * n; a++)
            for(int b = 0; b <= 2 * n; b++)
                for(int c = 0; c <= n; c++)
                    for(int d = 0; d <= n; d++)
                        ans = ans + (data)(g[a][b][c][d]) * (data)(f[n&1][a][b][c][d]);
        ans = ans + calzero();
    }
    print(ans);
    print(qpow(10, t1 + t2) - ans);
    return 0;
}

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#include <cmath>

#include <cstdio>

#include <vector>

#include <cstring>

#include <iostream>

#define inf 1000000000

#define rad 100000000

#define ll long long

using namespace std;

int n;

char str[100];

ll f[2][55][40][20][20], g[55][40][20][20]; // i 2 3 5 7 0?

struct data{

int l;ll v[20];

data(){

l=1;

memset(v,0,sizeof(v));

}

data(ll x){

memset(v,0,sizeof(v));

l=0;

while(x)

{

v[++l]=x%rad;

x/=rad;

}

ll& operator[](int x){

return v[x];

}

friend data operator+(data a,data b){

if(a.l<b.l)swap(a,b);

for(int i=1;i<=b.l;i++)

a[i]+=b[i];

for(int i=1;i<=a.l;i++)

if(a[i]>=rad)

{

if(i==a.l)a.l++;

a[i+1]+=a[i]/rad;

a[i]%=rad;

}

return a;

}

friend data operator-(data a,data b){

for(int i=1;i<=b.l;i++)

a[i]-=b[i];

for(int i=1;i<=a.l;i++)

if(a[i]<0)

{

a[i]+=rad;

a[i+1]--;

}

while(!a[a.l]&&a.l>1)a.l--;

return a;

}

friend data operator*(data a,ll b){

for(int i=1;i<=a.l;i++)

a[i]*=b;

for(int i=1;i<=a.l;i++)

if(a[i]>=rad)

{

a[i+1]+=a[i]/rad;

a[i]%=rad;

if(i==a.l)a.l++;

}

return a;

}

friend data operator*(data a,data b){

data c;

for(int i=1;i<=a.l;i++)

for(int j=1;j<=b.l;j++)

c[i+j-1]+=a[i]*b[j];

for(int i=1;i<=a.l+b.l;i++)

{

if(c[i]>=rad)

{

c[i+1]+=c[i]/rad;

c[i]%=rad;

}

if(c.v[i])c.l=i;

}

return c;

}

friend void print(data a){

printf("%lld",a[a.l]);

for(int i=a.l-1;i;i--)printf("%.8lld",a[i]);

puts("");

}

};

void dp(int L, int R)

{

memset(f, 0, sizeof(f));

int p = 0, q = 1;

f[0][0][0][0][0] = 1;

for(int x = L - 1; x <= R - 1; x++)

{

int l = x - L + 1;

char ch = str[x + 1];

for(int a = 0; a <= 3 * l; a++)

for(int b = 0; b <= 2 * l; b++)

for(int c = 0; c <= l; c++)

for(int d = 0; d <= l; d++)

if(f[p][a][b][c][d])

{

if(ch == '1' || ch == '?')

f[q][a][b][c][d] += f[p][a][b][c][d];

if(ch == '2' || ch == '?')

f[q][a+1][b][c][d] += f[p][a][b][c][d];

if(ch == '3' || ch == '?')

f[q][a][b+1][c][d] += f[p][a][b][c][d];

if(ch == '4' || ch == '?')

f[q][a+2][b][c][d] += f[p][a][b][c][d];

if(ch == '5' || ch == '?')

f[q][a][b][c+1][d] += f[p][a][b][c][d];

if(ch == '6' || ch == '?')

f[q][a+1][b+1][c][d] += f[p][a][b][c][d];

if(ch == '7' || ch == '?')

f[q][a][b][c][d+1] += f[p][a][b][c][d];

if(ch == '8' || ch == '?')

f[q][a+3][b][c][d] += f[p][a][b][c][d];

if(ch == '9' || ch == '?')

f[q][a][b+2][c][d] += f[p][a][b][c][d];

}

memset(f[p], 0, sizeof(f[p]));

swap(p, q);

}

data qpow(ll t, ll b)

{

data ans = data(1);

data a = (data)(t);

for(ll i = b; i; i >>= 1, a = a * a)

if(i & 1)ans = ans * a;

return ans;

}

data calzero()

{

int ans1 = 0, ans2 = 0;

for(int i = 1; i <= n; i++)

if(str[i] == '?')ans1++;

for(int i = n + 1; i <= 2 * n; i++)

if(str[i] == '?')ans2++;

data t1 = qpow(10, ans1) - qpow(9, ans1);

data t2 = qpow(10, ans2) - qpow(9, ans2);

return t1 * t2;

}

int main()

{

scanf("%d", &n);

scanf("%s", str+1);

ll t1 = 0, t2 = 0, zl = 0, zr = 0;

for(int i = 1; i <= 2 * n; i++)

if(str[i] == '?')

{

if(i <= n)t1++;

else t2++;

}

else if(str[i] == '0')

{

if(i <= n)zl = 1;

else zr = 1;

}

data ans;

if(zl || zr)

{

if(zl && zr)

ans = qpow(10, t1 + t2);

else if(zl)

ans = qpow(10, t1) * (qpow(10, t2) - qpow(9, t2));

else ans = (qpow(10, t1) - qpow(9, t1)) * qpow(10, t2);

}

else

{

dp(1, n);

memcpy(g, f[n&1], sizeof(f[n&1]));

dp(n + 1, 2 * n);

for(int a = 0; a <= 3 * n; a++)

for(int b = 0; b <= 2 * n; b++)

for(int c = 0; c <= n; c++)

for(int d = 0; d <= n; d++)

ans = ans + (data)(g[a][b][c][d]) * (data)(f[n&1][a][b][c][d]);

ans = ans + calzero();

}

print(ans);

print(qpow(10, t1 + t2) - ans);

return 0;

}

H:Stock Exchange POJ3903

#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define inf 1000000000
using namespace std;
int n;
int a[100005], f[100005];
int main()
{
        while(scanf("%d", &n) != EOF)
        {
                memset(f, 127, sizeof(f));
                for(int i = 1; i <= n; i++)
                        scanf("%d", &a[i]);
                for(int i = 1; i <= n; i++)
                        *lower_bound(f, f + n, a[i]) = a[i];
                printf("%d\n", lower_bound(f, f + n, inf) - f);
        }
        return 0;
}

#include <cmath>

#include <cstdio>

#include <vector>

#include <cstring>

#include <iostream>

#include <algorithm>

#define inf 1000000000

using namespace std;

int n;

int a[100005], f[100005];

int main()

{

while(scanf("%d", &n) != EOF)

{

memset(f, 127, sizeof(f));

for(int i = 1; i <= n; i++)

scanf("%d", &a[i]);

for(int i = 1; i <= n; i++)

*lower_bound(f, f + n, a[i]) = a[i];

printf("%d\n", lower_bound(f, f + n, inf) - f);

}

return 0;

}

« 「百度之星2017」程序设计大赛初赛(B) 【LOJ】小奇 NOIP 练习题 »

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算法设计与分析讨论班上机作业

凸包

A:Wall POJ1113

B:Scrambled Polygon POJ2007

动态规划 G 题真的坑

A:Fourier’s Lines POJ1923

B:Tour POJ2677

C:Increasing Sequences POJ1239

D:Charlie’s Change POJ1787

E:Strange Towers of Hanoi POJ1958

F:Balanced Lineup POJ3274

G:Banal Tickets POJ1608

H:Stock Exchange POJ3903

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