AIM Tech Round 4 (Div. 2) B. Rectangles
2017-08-25 08:17
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题目大意
求选出在同一行或同一列颜色相同的格子,共有多少种选法。题解
泥萌为什么都2^n什么的啊,我只会用组合数QAQ。杨辉三角预处理组合数,设b[i]为第i行1的个数,则:
ans+=∑i=1n∑j=1biCjbi
列同理,注意去掉块数为1的。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int read() { char ch=getchar();int f=0; while(ch<'0'||ch>'9') ch=getchar(); while(ch>='0'&&ch<='9') {f=f*10+(ch^48);ch=getchar();} return f; } long long c[55][55],ans;long long a[55][55],b[55],d[55]; int main() { int n=read(),m=read(); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { a[i][j]=read(); if(a[i][j]) { b[i]++; d[j]++; } } } for(int i=0;i<=50;i++) { c[i][0]=1; for(int j=1;j<=i;j++) { c[i][j]=c[i-1][j-1]+c[i-1][j]; } } for(int i=1;i<=n;i++) { for(int j=1;j<=b[i];j++) { ans+=c[b[i]][j]; } for(int j=1;j<=m-b[i];j++) { ans+=c[m-b[i]][j]; } } for(int i=1;i<=m;i++) { for(int j=2;j<=d[i];j++) { ans+=c[d[i]][j]; } for(int j=2;j<=n-d[i];j++) { ans+=c[n-d[i]][j]; } } cout<<ans; }
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