「CF453B」Little Pony and Harmony Chest
Princess Twilight went to Celestia and Luna’s old castle to research the chest from the Elements of Harmony.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of elements of the sequences a and b. The next line contains n integers a1, a2, …, an (1 ≤ ai ≤ 30).
Output
Output the key — sequence bi that minimizes the sum described above. If there are multiple optimal sequences, you can output any of them.
Sample test(s)
input
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1 2 |
5 1 1 1 1 1 |
output
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1 |
1 1 1 1 1 |
input
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1 2 |
5 1 6 4 2 8 |
output
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1 |
1 5 3 1 8 |
题解
一开始以为是相邻互质。。。后来发现过不了样例T T
如果相邻互质直接dp即可
大概这样
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#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #define ll long long #define inf 1000000000 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,ans,mn=inf,v[100005]; int f[105][45],fa[105][45],q[105]; inline int gcd(int a,int b) { return b==0?a:gcd(b,a%b); } int main() { memset(f,127/3,sizeof(f)); n=read(); for(int i=1;i<=n;i++) v[i]=read(); for(int i=1;i<=40;i++)f[1][i]=abs(v[1]-i); for(int i=2;i<=n;i++) for(int a=1;a<=40;a++) { for(int b=1;b<=40;b++) if(gcd(a,b)==1) { if(f[i-1][b]<f[i][a]) { f[i][a]=f[i-1][b];fa[i][a]=b; } } f[i][a]+=abs(v[i]-a); } for(int i=1;i<=40;i++) if(f[n][i]<mn) { ans=i;mn=f[n][i]; } for(int i=n;i;i--) { q[i]=ans;ans=fa[i][ans]; } for(int i=1;i<=n;i++) printf("%d",q[i]); return 0; } |
两两互质就比较麻烦了
首先1-60的质数有17个，去掉59，因为取59不如取1
那么就剩下16个，状压一下T T，转移枚举使用了那个数字，应该要预处理下每个数字用了哪些质因数吧T T，不然似乎比较虚
复杂度2^16*100*60是三亿多，在cf的机子下并且非常多废状态还是妥妥过的啦
我最大的点似乎只跑一秒左右。。。时限是4秒啊
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#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #define ll long long #define inf 1000000000 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 bin[20]; int n,cnt,ed,mn=inf,ans=0;; int f[105][65536],fa[105][65536]; int v[105],q[105],a[60]; int p[50]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53}; inline int cal(int x) { int sum=0; for(int i=0;i<16;i++) if(x%p[i]==0)sum+=bin[i]; return sum; } int main() { bin[0]=1;for(int i=1;i<20;i++)bin[i]=bin[i-1]<<1; for(int i=1;i<59;i++)a[i]=cal(i); memset(f,-1,sizeof(f)); n=read(); for(int i=1;i<=n;i++)v[i]=read(); ed=(1<<16)-1; f[0][0]=0; for(int i=0;i<n;i++) for(int j=0;j<=ed;j++) if(f[i][j]!=-1) for(int k=1;k<59;k++) { int to=j+a[k],t=f[i][j]+abs(k-v[i+1]); if((j|a[k])==to&&(t<f[i+1][to]||f[i+1][to]==-1)) { f[i+1][to]=t; fa[i+1][to]=k; } } for(int i=0;i<=ed;i++) if(f[n][i]<mn&&f[n][i]!=-1) { ans=i;mn=f[n][i]; } for(int i=n;i;i--) { q[i]=fa[i][ans];ans-=a[q[i]]; } for(int i=1;i<=n;i++) printf("%d ",q[i]); return 0; } |
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