Four Operations

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38 Accepted Submission(s): 18

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

1
12345

Sample Output

Case #1: 1

Source

2016年中国大学生程序设计竞赛（杭州）

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/*

枚举减号，刚开始天真的以为，除数最多是两位......卡死。

*/

#include<bits/stdc++.h>

#define ll long long

#define INF 0x3fffffffffffffff

#define N 22

using namespace std;

string op;

ll right(string s)

{

    ll e=;

    int n=s.size();

    for(int i=;i<n;i++)

        e=e*+s[i]-'';

    //cout<<c<<" "<<d<<" "<<e<<endl;

    ll ans=(s[]-'')*(s[]-'')/e;

    return ans;

}

ll left(string s)//减号左边的部分

{

    ll cur1=,cur2=;

    int n=s.size();

    for(int i=;i<n;i++)

        cur1=cur1*+(s[i]-'');

    cur1+=s[]-'';

    for(int i=;i<n-;i++)

        cur2=cur2*+(s[i]-'');

    cur2+=s[n-]-'';

    //cout<<"max(cur1,cur2)="<<max(cur1,cur2)<<" ";

    return max(cur1,cur2);

}

ll solve(string s)

{

    ///枚举减号

    ll cur=-INF,s1,s2;

    int n=s.size();

    for(int i=;i<=n-;i++)

    {

        s1=left(s.substr(,i));

        s2=right(s.substr(i,n-i));

        //cout<<s1<<" "<<s2<<endl;

        cur=max(cur,s1-s2);

    }

    printf("%lld\n",cur);

}

int t;

int main()

{

    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);

    scanf("%d",&t);

    for(int Case=;Case<=t;Case++)

    {

        printf("Case #%d: ",Case);

        cin>>op;

        solve(op);

    }

    return ;

}

巴特西

2016年中国大学生程序设计竞赛（杭州）1006 Four Operations

Four Operations

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