HDU-5938：Four Operations（DP）

Four Operations

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2082    Accepted Submission(s): 608

Problem Description

Little Ruins is a studious boy, recently he learned the four operations!

Now he want to use four operations to generate a number, he takes a string which only contains digits '1' - '9', and split it into 5 intervals and add the four operations '+', '-', '*' and '/' in order, then calculate the result(/ used as integer division).

Now please help him to get the largest result.

 

Input

First line contains an integer T, which indicates the number of test cases.

Every test contains one line with a string only contains digits '1'-'9'.

Limits
1≤T≤105
5≤length of string≤20

 

Output

For every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the result.

 

Sample Input

112345

 

Sample Output

Case #1: 1

思路：d[i]表示[0,i]间放'+'的最大值，f[i]表示[i,n]间放'*'和'/'后的最小值。ans=max(d[i]-f[i+1])；

#include<bits/stdc++.h>using namespace std;char s[30];long long d[30],f[30],n;long long num(int x,int y){    long long sum=0;    for(int i=x;i<=y;i++)sum=sum*10+s[i]-'0';    return sum;}int main(){    int T,cas=1;cin>>T;    while(T--)    {        memset(d,0,sizeof d);        memset(f,1e9+7,sizeof f);        scanf("%s",s);        n=strlen(s);        for(int i=0;i<n-3;i++)        {            for(int j=0;j<i;j++)d[i]=max(d[i],num(0,j)+num(j+1,i));        }        for(int i=n-1;i>=2;i--)        {            for(int j=i;j<=n-1;j++)            {                for(int k=j+1;k<n-1;k++)f[i]=min(f[i],num(i,j)*num(j+1,k)/num(k+1,n-1));            }        }        long long ans=-1e9-7;        for(int i=1;i<n-3;i++)ans=max(ans,d[i]-f[i+1]);        printf("Case #%d: %lld\n",cas++,ans);    }    return 0;}