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}i
th
room, there is a wizard who has a resentment value of a[i]a[i]. The prince has MM curses, the j^{th}j
th
curse is f[j]f[j], and f[j]f[j] represents one of the four arithmetic operations, namely addition(‘+’), subtraction(‘-‘), multiplication(‘*’), and integer division(‘/’). The prince’s initial resentment value is KK. Entering a room and fighting with the wizard will eliminate a curse, but the prince’s resentment value will become the result of the arithmetic operation f[j]f[j] with the wizard’s resentment value. That is, if the prince eliminates the j^{th}j
th
curse in the i^{th}i
th
room, then his resentment value will change from xx to (x\ f[j]\ a[i]x f[j] a[i]), for example, when x=1, a[i]=2, f[j]=x=1,a[i]=2,f[j]=’+’, then xx will become 1+2=31+2=3.

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],…,aN, 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.

样例输入 复制
3
2 1 5
2 3
/
3 2 1
1 2 3
++
4 4 5
1 2 3 4
+-*/
样例输出 复制
2
6
3
题目链接：https://nanti.jisuanke.com/t/31711
题意：
t组样例，n,m,k接下来有n个数，m次操作，初值为k
按顺序执行+或-或*或/，问你执行了m次操作之后k的最大值是多少
题解：
那么我们只要开二维的dp数组，dpbig[i][j]代表到第i个数执行了j次操作最大的值是多少
但是要注意这里的*可能是乘负数会更优，比如说10 10 -1000 -1000 10 10 10 的时候，乘上2个-1000肯定比其他的乘起来要大。所以我们还需要开一个dpsmall数组记录最小值，那么算dpbig的时候就有一种可能是从dpsmall里乘一个负数得到最大的答案。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=(int)1e9+7;
const int maxn=1005;
ll a[maxn],dpbig[maxn][10],dpsmall[maxn][10];
char op[maxn];
int n,m,k;
ll deal(ll x,char op,ll y){
    if(op=='+') return x+y;
    if(op=='-') return x-y;
    if(op=='*') return x*y;
    if(op=='/') return x/y;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
        }
        scanf("%s",op+1);
        for(int i=0;i<=n;i++){
            for(int j=0;j<=m;j++){
                dpbig[i][j]=-1e18-5;
            }
        }
        for(int i=0;i<=n;i++){
            for(int j=0;j<=m;j++){
                dpsmall[i][j]=1e18+5;
            }
        }
        for(int i=0;i<=n;i++){
            dpbig[i][0]=dpsmall[i][0]=k;
        }

        for(int i=1;i<=n;i++){
            for(int j=1;j<=min(i,m);j++){
                if(a[i]>0||op[j]=='-'||op[j]=='+'){
                    ll now=deal(dpbig[i-1][j-1],op[j],a[i]);
                    dpbig[i][j]=max(dpbig[i][j],max(dpbig[i-1][j],now));
                    now=deal(dpsmall[i-1][j-1],op[j],a[i]);
                    dpsmall[i][j]=min(dpsmall[i][j],min(dpsmall[i-1][j],now));
                }
                else {
                    ll now=deal(dpbig[i-1][j-1],op[j],a[i]);
                    dpsmall[i][j]=min(dpsmall[i][j],min(dpsmall[i-1][j],now));
                    now=deal(dpsmall[i-1][j-1],op[j],a[i]);
                    dpbig[i][j]=max(dpbig[i][j],max(dpbig[i-1][j],now));
                }
            }
        }
        ll ans=-1e18-5;
        for(int i=1;i<=n;i++){
            ans=dpbig[i][m];
        }
        printf("%lld\n",ans);
    }
    return 0;
}