spoj 345

DP  想了好久  还是看了一下题解。。。。 f[i][j]表示i到j全部合并后的最小花费,f[i][j] = min{f[i][k]+f[k+1][j]+d[i][k]*d[k+1][j]} (i <= k <= j-1)

#include <iostream>
#include <fstream>
#include <cstdio>
#include <cstring>
using namespace std;

#define INF 100000000

int f[101][101], d[101];
int n;

int main()
{
    while (scanf("%d",&n) == 1)
    {
        memset(d, 0, sizeof(d));
        memset(f, 0, sizeof(f));
        for (int i = 1; i <= n; i++)
            scanf("%d",&d[i]), d[i] += d[i-1];
        for (int k = 2; k <= n; k++)
            for (int i = 1; i <= n-k+1; i++)
            {
                int j = i+k-1;
                f[i][j] = INF;
                for (int q = i; q <= j-1; q++)
                    f[i][j] = min(f[i][j], f[i][q]+f[q+1][j]+((d[q]-d[i-1])%100)*((d[j]-d[q])%100));
            }
        printf("%d\n",f[1][n]);
    }
    return 0;
}