A prince of the Science Continent was imprisoned in a castle because of his contempt for mathematics when he was young, and was entangled in some mathematical curses. He studied hard until he reached adulthood and decided to use his knowledge to escape the castle.

There are NN rooms from the place where he was imprisoned to the exit of the castle. In the i^{th}ith room, there is a wizard who has a resentment value of a[i]a[i]. The prince has MM curses, the j^{th}jth curse is f[j]f[j], and f[j]f[j] represents one of the four arithmetic operations, namely addition(‘+‘

Before the prince escapes from the castle, he must eliminate all the curses. He must go from a[1]a[1] to a[N]a[N] in order and cannot turn back. He must also eliminate the f[1]f[1] to f[M]f[M] curses in order(It is guaranteed that N\ge MN≥M). What is the maximum resentment value that the prince may have when he leaves the castle?

Input

The first line contains an integer T(1 \le T \le 1000)T(1≤T≤1000), which is the number of test cases.

For each test case, the first line contains three non-zero integers: N(1 \le N \le 1000), M(1 \le M \le 5)N(1≤N≤1000),M(1≤M≤5) and K(-1000 \le K \le 1000K(?1000≤K≤1000), the second line contains NN non-zero integers: a[1], a[2], ..., a[N](-1000 \le a[i] \le 1000)a[1],a[2],...,a[N](?1000≤a[i]≤1000), and the third line contains MM characters: f[1], f[2], ..., f[M](f[j] =f[1],f[2],...,f[M](f[j]=‘+‘,‘-‘,‘*‘,‘/‘, with no spaces in between.

Output

For each test case, output one line containing a single integer.

樣例輸入復制

3
2 1 5
2 3
/
3 2 1
1 2 3
++
4 4 5
1 2 3 4
+-*/

樣例輸出復制

2
6
3

題目來源

ACM-ICPC 2018 焦作賽區網絡預賽

題解：DP；

分別記錄最大值和最小值，（因為可能出現兩個都是負數的情況），轉移方程為：

dp[i][j] = dp[i - 1][j];  dp1[i][j] = dp1[i - 1][j];分別記錄最大和最小
對於不同符號，有不同的轉移方程

參考代碼：

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <string>
#include <algorithm>
#include <bitset>
#define INF 0x3f3f3f3f3f3f3f3fll
#define clr(x, y) memset(x, y, sizeof(x))
#define mod 1000000007
using namespace std;
typedef long long LL;
const int maxn = 1010;
const int maxm = 10;
int a[maxn],t;
LL dp[maxn][maxm],dp1[maxn][maxn];
char f[maxm];
LL Max(LL a,LL b) {return a>=b? a:b; }
LL Min(LL a,LL b)  { return a<b? a:b; }
int main()
{
    scanf("%d", &t);
    while(t--)
    {
        memset(dp, -INF, sizeof dp);
        memset(dp1,INF,sizeof dp1);
        int n, m, k;
        scanf("%d%d%d", &n, &m, &k);
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            dp[i][0] =dp1[i][0]=k;
        }
        dp[0][0] = dp1[0][0] = k;
        scanf("%s", f + 1);
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
            	if(i<j) continue;
                dp[i][j] = dp[i - 1][j]; dp1[i][j] = dp1[i - 1][j];
                if(f[j] == ‘+‘) dp[i][j]=Max(dp[i][j],dp[i - 1][j - 1] + a[i]), dp1[i][j]=Min(dp1[i][j],dp1[i - 1][j - 1] + a[i]);
                else if(f[j] == ‘-‘) dp[i][j] = Max(dp[i][j], dp[i - 1][j - 1] - a[i]), dp1[i][j] = Min(dp1[i][j], dp1[i - 1][j - 1] - a[i]);
                else if(f[j] == ‘*‘)
				{
					dp[i][j] = Max(dp[i][j], Max(dp[i - 1][j - 1] * a[i],dp1[i - 1][j - 1] * a[i]) );
					dp1[i][j] = Min(dp1[i][j], Min(dp[i - 1][j - 1] * a[i],dp1[i - 1][j - 1] * a[i]) );
				} 
                else 
				{
					dp[i][j] = Max(dp[i][j], Max(dp[i - 1][j - 1] / a[i], dp1[i - 1][j - 1] / a[i]) );
					dp1[i][j]=Min(dp1[i][j], Min(dp[i - 1][j - 1] / a[i], dp1[i - 1][j - 1] / a[i]) );
				}
            }
        }
        printf("%lld\n", Max(dp[n][m],dp1[n][m]) );
    }
    return 0;
}

ACM-ICPC 2018 焦作賽區網絡預賽 B題 Mathematical Curse