部落格目錄

原題

題目傳送門

•  22.08%
•  1000ms
•  65536K

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('+'

), 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}jth curse in the i^{th}ith 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],...,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.

樣例輸入複製

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

樣例輸出複製

2
6
3

題目來源

思路

動態規劃，設狀態dp[i][j]表示取前i個數，j個字元時能獲得的最值（最大和最小值都要維護）

設mul(dp,i,j)表示之前的結果為dp，第i個數字第j個運算子運算後的結果

有轉移方程：

dpmax[i][j]=max(dpmax[i-1][j]，mul(dpmax[i-1][j-1],i,j)，mul(dpmin[i-1][j-1],i,j))

//別忘了負數乘以負數可能會轉移到最大值。最小值同理。

dpmin同上（詳見程式碼）

邊界自然是

dpmax=-inf

dpmin=inf

dp[i][0]=k

AC程式碼

#include<bits/stdc++.h> using namespace std; typedef long long ll; ll dpmin[1005][10]; ll dpmax[1005][10]; ll const inf=0x3f3f3f3f3f3f;

int numb[1005]; char s[10]; inline ll mul(ll v,int i,int j){     switch(s[j]){         case '+':return v+numb[i];         case '-':return v-numb[i];         case '*':return v*numb[i];         case '/':return v/numb[i];     }     return 0; } int main(){     int T;     cin>>T;     int n,m,k;     while(T--){         scanf("%d%d%d",&n,&m,&k);         for(int i=1;i<=n;i++){             scanf("%d",numb+i);         }         for(int i=0;i<=n;i++){             for(int j=0;j<=m;j++){                 dpmax[i][j]=-inf;                 dpmin[i][j]=inf;             }         }         dpmin[0][0]=dpmax[0][0]=k;         getchar();         gets(s+1);         for(int i=1;i<=n;i++){             dpmin[i][0]=k;             dpmax[i][0]=k;             for(int j=1;j<=m;j++){                 dpmin[i][j]=dpmin[i-1][j];                  dpmax[i][j]=dpmax[i-1][j];                  if(dpmax[i-1][j-1]!=-inf)                     dpmax[i][j]=max(dpmax[i][j],mul(dpmax[i-1][j-1],i,j));                 if(dpmin[i-1][j-1]!=inf)                     dpmax[i][j]=max(dpmax[i][j],mul(dpmin[i-1][j-1],i,j));                              if(dpmin[i-1][j-1]!=inf)                     dpmin[i][j]=min(dpmin[i][j],mul(dpmin[i-1][j-1],i,j));                     if(dpmax[i-1][j-1]!=-inf)                     dpmin[i][j]=min(dpmin[i][j],mul(dpmax[i-1][j-1],i,j));                          }         }         cout<<dpmax[n][m]<<endl;     } }