1274. Pascal's Travels
2012-04-11 22:06
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dp[i][j]表示到从左上角(1,1)到(i,j)的方案数 dp[i][j]+=(dp[i][u])+(dp[v][j]),其中 1<=u<j, 1<=v<i, (i,u)能走到(i,j), (v,j)能走到(i,j)
#include<iostream> //DP #include <cstdio> #include<cstring> using namespace std; char g[40][40]; long long ans[40][40]; //题目要求是64位 int main() { int n; while(cin>>n,n!=-1) { for(int i=1;i<=n;++i) scanf("%s",g[i]+1); memset(ans,0,sizeof(ans)); ans[1][1]=1; for(int i=1;i<=n;++i) for(int j=1;j<=n;++j) { for(int u=1;u<j;++u) if(g[i][u]=='0'+j-u) ans[i][j]+=ans[i][u]; for(int v=1;v<i;++v) if(g[v][j]=='0'+i-v) ans[i][j]+=ans[v][j]; } cout<<ans <<endl; } return 0; }
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