hdu 2751 命运

/* Name: hdu 2751 命运 Author:UnimenSun Date: 30/06/11 21:54 Description:简单DP *//*解题报告：简单DP转移方程： dp[i][j] = max(dp[i-1][j], dp[i][j-1], dp[i][k]) + array[i][j] k的条件：1<=k<j && j%k==0 其中dp[i][j]表示到i, j位置的最大值 */#include <iostream>#include <cstring>#include <algorithm>using namespace std;int c;int n, m;int dp[25][1010]; int main(){int i, j, k;int nTemp;while(cin>>c){while(c--){cin>>n>>m;//初始化 for(i=0; i<=n; ++i)for(j=0; j<=m; ++j)dp[i][j] = -9999999;dp[1][0] = dp[0][1] = 0;for(i=1; i<=n; ++i)for(j=1; j<=m; ++j)cin>>dp[i][j];for(i=1; i<=n; ++i)for(j=1; j<=m; ++j){nTemp = dp[i-1][j];nTemp = max(nTemp, dp[i][j-1]);for(k=1; k<j; ++k){if(j%k == 0)nTemp = max(nTemp, dp[i][k]);}dp[i][j] = nTemp + dp[i][j];}cout<<dp[n][m]<<endl;}}return 0;}