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

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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liuyiding

/*
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 ;
}