CODEVS-1039 划分DP 解法类似最佳加法表达式

中文题 不贴题意了

直接上代码

code

	#include<iostream>
	#include<cstdio>
	#include<cstdlib>
	#include<algorithm>
	#include<cmath>
	#include<string>
	#include<cstring>
	#pragma warning(disable:4996)

	using namespace std;
	typedef long long ll;
	const int MAX = 250;
	const int INF = 0x7fffff;
	int sum[MAX][MAX];
	int w[MAX];
	int pre[MAX][MAX];//前缀数组
	ll dp[MAX][MAX];
	int n,k;
	int main()
	{
		scanf("%d%d",&n,&k);
		char s[50];
		scanf("%s",s);
		memset(pre, 0, sizeof(pre));
		for (int i = 1; i <= n; i++)
		{
			w[i] = s[i-1] - '0';
			pre[i][i] = w[i];
		}
		for (int i = 1; i <= n; i++)
		{
			for (int j = i + 1; j<=n; j++)
			{
				
					pre[i][j] = pre[i][j - 1] * 10 + pre[j][j];

			}
		}
		for (int i = 1; i <= n; i++)
			dp[i][0] = pre[1][i];
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= k; j++)
			{
				if
					( j >= i)break;
				else
				{
					for (int x = 1; x < i; x++)
					{
						dp[i][j] = max(dp[i][j], dp[x][j - 1] * pre[x + 1][i]);
					}
				}
			}
		}
		printf("%d\n",dp[n][k]);
		system("pause");
	}

状态转移方程是在前i个数插入k个乘号 （需要预处理一下）=前j个数插入k+1个乘号并乘上j+1-i这些数字所组成的数