「CF497B」Tennis Game

2014年12月18日3,6070

Petya and Gena love playing table tennis. A single match is played according to the following rules: a match consists of multiple sets, each set consists of multiple serves. Each serve is won by one of the players, this player scores one point. As soon as one of the players scores t points, he wins the set; then the next set starts and scores of both players are being set to 0. As soon as one of the players wins the total of s sets, he wins the match and the match is over. Here s and t are some positive integer numbers.

To spice it up, Petya and Gena choose new numbers s and t before every match. Besides, for the sake of history they keep a record of each match: that is, for each serve they write down the winner. Serve winners are recorded in the chronological order. In a record the set is over as soon as one of the players scores t points and the match is over as soon as one of the players wins s sets.

Petya and Gena have found a record of an old match. Unfortunately, the sequence of serves in the record isn’t divided into sets and numbers s and t for the given match are also lost. The players now wonder what values of s and t might be. Can you determine all the possible options?

Input

The first line contains a single integer n — the length of the sequence of games (1 ≤ n ≤ 10⁵).

The second line contains n space-separated integers a_i. If a_i = 1, then the i-th serve was won by Petya, if a_i = 2, then the i-th serve was won by Gena.

It is not guaranteed that at least one option for numbers s and t corresponds to the given record.

Output

In the first line print a single number k — the number of options for numbers s and t.

In each of the following k lines print two integers s_i and t_i — the option for numbers s and t. Print the options in the order of increasing s_i, and for equal s_i — in the order of increasing t_i.

Sample test(s)

input

5
1 2 1 2 1

1 2	5 1 2 1 2 1

output

2
1 3
3 1

1 3

3 1

input

4
1 1 1 1

1 1 1 1

output

1 4

2 2

4 1

input

4
1 2 1 2

1 2 1 2

output

input

8
2 1 2 1 1 1 1 1

1 2	8 2 1 2 1 1 1 1 1

output

1 6

2 3

6 1

题解

我一开始写成了暴力TAT

获胜的一定是最后一个，设为x，另一人y

枚举T，对于每一场i，记录下次x的得分点nxt[i]。。。

暴力可以每次从now移到nxt[now]（即让x得1分），判y加上now到nxt[now]之间的得分数是否超过T，若超过，则让y得分到T为止。。。

改用二分，前缀和或者倍增每次x让得T分就行了。。。

复杂度nlogn或者nlognlogn

测了下暴力 TLE23

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<set>
#include<ctime>
#include<vector>
#include<queue>
#include<algorithm>
#include<map>
#define inf 1000000000
#define pa pair<int,int>
#define ll long long 
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,K,top;
int a[100005],nxt[100005];
struct data{
    int t,s;
}ans[100005];
bool operator<(data a,data b)
{
    if(a.s==b.s)return a.t<b.t;
    return a.s<b.s;
}
int main()
{
    n=read();
    for(int i=1;i<=n;i++)a[i]=read();
    int win=a[n];
    int last=0;
    for(int i=n;i;i--)
    {
        nxt[i]=last;
        if(a[i]==win)last=i;
    }
    nxt[0]=last;
    for(int T=1;T<=n;T++)
    {
        int t1=0,t2=0,w1=0,w2=0,now=0;
        while(now<n)
        {
            int tmp=T-t2;
            if(nxt[now]-now-1<tmp)
            {
                t2+=nxt[now]-now-1;t1++;
                now=nxt[now];
                if(t1==T)w1++,t1=t2=0;
            }
            else now+=tmp,w2++,t1=t2=0;
        }
        if(t1!=0||w2>=w1)continue;
        K++;
        ans[K].t=T;ans[K].s=w1;
    }
    sort(ans+1,ans+K+1);
    printf("%d\n",K);
    for(int i=1;i<=K;i++)
        printf("%d %d\n",ans[i].s,ans[i].t);
    return 0;
}

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<set>

#include<ctime>

#include<vector>

#include<queue>

#include<algorithm>

#include<map>

#define inf 1000000000

#define pa pair<int,int>

#define ll long long

using namespace std;

inline int read()

{

int x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

int n,K,top;

int a[100005],nxt[100005];

struct data{

int t,s;

}ans[100005];

bool operator<(data a,data b)

{

if(a.s==b.s)return a.t<b.t;

return a.s<b.s;

}

int main()

{

n=read();

for(int i=1;i<=n;i++)a[i]=read();

int win=a[n];

int last=0;

for(int i=n;i;i--)

{

nxt[i]=last;

if(a[i]==win)last=i;

}

nxt[0]=last;

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

{

int t1=0,t2=0,w1=0,w2=0,now=0;

while(now<n)

{

int tmp=T-t2;

if(nxt[now]-now-1<tmp)

{

t2+=nxt[now]-now-1;t1++;

now=nxt[now];

if(t1==T)w1++,t1=t2=0;

}

else now+=tmp,w2++,t1=t2=0;

}

if(t1!=0||w2>=w1)continue;

K++;

ans[K].t=T;ans[K].s=w1;

}

sort(ans+1,ans+K+1);

printf("%d\n",K);

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

printf("%d %d\n",ans[i].s,ans[i].t);

return 0;

}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<set>
#include<ctime>
#include<vector>
#include<queue>
#include<algorithm>
#include<map>
#define inf 1000000000
#define pa pair<int,int>
#define ll long long 
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,K,top;
int a[100005],sum1[100005],sum2[100005];
struct data{
    int t,s;
}ans[100005];
bool operator<(data a,data b)
{
    if(a.s==b.s)return a.t<b.t;
    return a.s<b.s;
}
int main()
{
    n=read();
    for(int i=1;i<=n;i++)a[i]=read();
    int flag=a[n];
    for(int i=1;i<=n;i++)
    {
        sum1[i]=sum1[i-1]+(a[i]==flag);
        sum2[i]=sum2[i-1]+(a[i]!=flag);
    }
    for(int T=1;T<=n;T++)
    {
        int t1=0,t2=0,w1=0,w2=0,now=0;
        while(now<n)
        {
            if(lower_bound(sum1+now+1,sum1+n+1,sum1[now]+(T-t1))==sum1+n+1)break;
            int nxt=lower_bound(sum1+now+1,sum1+n+1,sum1[now]+(T-t1))-sum1;
            int tmp=T-t2;
            if(nxt-now-(T-t1)<tmp)
            {
                now=nxt;
                t1=t2=0;
                w1++;
            }
            else now=lower_bound(sum2+now+1,sum2+n+1,sum2[now]+tmp)-sum2,w2++,t1=t2=0;
        }
        if(t1!=0||w2>=w1||now!=n)continue;
        K++;
        ans[K].t=T;ans[K].s=w1;
    }
    sort(ans+1,ans+K+1);
    printf("%d\n",K);
    for(int i=1;i<=K;i++)
        printf("%d %d\n",ans[i].s,ans[i].t);
    return 0;
}

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<set>

#include<ctime>

#include<vector>

#include<queue>

#include<algorithm>

#include<map>

#define inf 1000000000

#define pa pair<int,int>

#define ll long long

using namespace std;

inline int read()

{

int x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

int n,K,top;

int a[100005],sum1[100005],sum2[100005];

struct data{

int t,s;

}ans[100005];

bool operator<(data a,data b)

{

if(a.s==b.s)return a.t<b.t;

return a.s<b.s;

}

int main()

{

n=read();

for(int i=1;i<=n;i++)a[i]=read();

int flag=a[n];

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

{

sum1[i]=sum1[i-1]+(a[i]==flag);

sum2[i]=sum2[i-1]+(a[i]!=flag);

}

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

{

int t1=0,t2=0,w1=0,w2=0,now=0;

while(now<n)

{

if(lower_bound(sum1+now+1,sum1+n+1,sum1[now]+(T-t1))==sum1+n+1)break;

int nxt=lower_bound(sum1+now+1,sum1+n+1,sum1[now]+(T-t1))-sum1;

int tmp=T-t2;

if(nxt-now-(T-t1)<tmp)

{

now=nxt;

t1=t2=0;

w1++;

}

else now=lower_bound(sum2+now+1,sum2+n+1,sum2[now]+tmp)-sum2,w2++,t1=t2=0;

}

if(t1!=0||w2>=w1||now!=n)continue;

K++;

ans[K].t=T;ans[K].s=w1;

}

sort(ans+1,ans+K+1);

printf("%d\n",K);

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

printf("%d %d\n",ans[i].s,ans[i].t);

return 0;

}

« 「BZOJ1822」[JSOI2010] Frozen Nova 冷冻波「BZOJ2282」[SDOI2011] 消防 »

hzwer.com _ Home | http://hzwer.com

「CF497B」Tennis Game

分类目录

热门文章

Myfriends

近期评论