划分型动态规划 之 CODE[VS] 1017 乘积最大 2000年NOIP全国联赛提高组

/*
dp[i][k] := 截取并获得原字符串前面长度为i+1的字符串，使用K个乘号将它分成K+1个部分，这K+1个部分的乘积的最大值
初始化：
	dp[][] = { 0 }
	dp[i][0] = Num(0, i); // 截取并获得原字符串前面长度为(i+1)的字符串（字符串0~i位置），返回其所对应的数值
状态方程：
	dp[i][k] = max(dp[i][k], dp[j][k-1]*Num(j+1, i))
	0<=j<i
答案：
	dp[N-1][K]
*/

#define _CRTDBG_MAP_ALLOC
#include <stdlib.h>
#include <crtdbg.h>
#define _CRT_SECURE_NO_WARNINGS
#define HOME

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstddef>
#include <iterator>
#include <algorithm>
#include <string>
#include <locale>
#include <cmath>
#include <vector>
#include <cstring>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MaxN = 45;
const int MaxK = 10;


int N, K;
string numStr;
int dp[MaxN][MaxK] = { 0 };


int Num(int ibegin, int iend)
{
    int num = 0;
    int tenMul = 1;
    for (int i = iend; i >= ibegin; --i)
    {
        num += ((numStr[i] - '0') * tenMul);
        tenMul *= 10;
    }
    return num;
}

void Solve()
{
    for (int i = 0; i < N; ++i)
    {
        for (int j = 0; j < i; ++j)
        {
            for (int k = 1; k <= K; ++k)
            {
                //printf("%d -> %d : %d
", j+1, i, Num(j + 1, i));
                dp[i][k] = max(dp[i][k], dp[j][k - 1] * Num(j + 1, i));
                //printf("    %d -> %d : %d
", j, k-1, dp[j][k-1]);
                //printf("    %d -> %d : %d
", i, k, dp[i][k]);
            }
        }
    }
    cout << dp[N - 1][K] << endl;
}

int main() 
{
#ifdef HOME
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
#endif

    cin >> N >> K;
    cin >> numStr;
    for (int i = 0; i < N; ++i)
    {
        dp[i][0] = Num(0, i);
    }
    Solve();

#ifdef HOME
    cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
    _CrtDumpMemoryLeaks();
    system("pause");
#endif
    return 0;
}