[wikioi]乘积最大

http://wikioi.com/problem/1017/

划分型动态规划
1.转移方程是:f[i][j]=max(f[k][j-1]*t[k+1][i]),f[i][j]表示前面i个字符加上j个乘号所得的最大值,t[i][j]表示i到j的数值;
2.可预处理,先计算出每段的数字值;
3.看代码分析错误实在不行时,还是debug一下吧。

#include <cstring>
#include <iostream>
#define ulong long long
using namespace std;
ulong t[45][45]; // number from i to j; start from 0;
ulong f[45][10]; // the first i digits, seperated to  j part, max value;
int data[45] = {0};

int main()
{
    int n, k;
    cin >> n >> k;
    memset(t, 0, sizeof(t));
    memset(f, 0, sizeof(f));
    // input
    char c;
    for (int i = 0; i < n; i++)
    {
        cin >> c;
        data[i] = c - '0';
        t[i][i] = data[i];
    }
    // pre-process
    for (int i = 0; i < n; i++)
    {
        for (int j = i+1; j < n; j++)
        {
            t[i][j] = t[i][j-1]*10 + data[j];
        }
    }
    // dp
    for (int j = 0; j <=k ; j++)
    {
        for (int i = j+1; i <= n; i++)
        {
            if (j == 0)
            {
                f[i][0] = t[0][i-1];
            }
            else
            {
                ulong max = 0;
                for (int x = 1; x < i; x++)
                {
                    ulong tmp = f[x][j-1] * t[x][i-1];
                    if (tmp > max) max = tmp;
                }
                f[i][j] = max;
            }
        }
    }
    cout << f[n][k] << endl;
    return 0;
}

  

posted @ 2013-09-20 12:18  阿牧遥  阅读(156)  评论(0编辑  收藏  举报