uva 10891 - Game of Sum

题解   白书上  总和一定的  dp 题；经典题

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
using namespace std;

int dp[105][105],M,sum[105],arr[105];
int main( )
{
    while( scanf("%d",&M) && M )
    {
        memset( dp,0,sizeof(dp) );
        memset( sum,0,sizeof(sum) );
        for( int i = 1; i <= M; i++ )
        {
            scanf("%d",&arr[i]);
            sum[i] = sum[i-1]+arr[i];
        }
           for( int i = 0; i <= M; i++ )dp[i][i] = arr[i];
        for( int i = 1; i < M; i++ )
        for( int j = 1; j+i <= M; j++ )
        {
            int l = i+j; int ans = sum[l] - sum[j-1];
            int Min = 0;
            for( int k = j; k <= l; k++ )
            {
                Min = min( Min,dp[j][k] );
                Min = min( Min,dp[k][l] );
            }
            dp[j][l] = ans - Min;
        }
        cout<<dp[1][M]*2-sum[M]<<endl;
    }
    return 0;
}