2016ccpc杭州赛 hdu 5938 F.Four Operations

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本文介绍了一道编程竞赛题目，任务是通过四则运算使给定数字串生成的最大值。文章提供了完整的代码实现和解题思路，指出关键在于合理分割数字串并选择运算符。

Four Operations

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 79 Accepted Submission(s): 38

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 | We have carefully selected several similar problems for you: 5943 5942 5941 5940 5939

Statistic | Submit | Discuss | Note 题意：给你一个字符串，把它分成四块，按顺序把符号+ - * /放上去，问能得到的最大数是多少。

思路：又是一道水题。首先我们看*和/，因为前面有-，所以后面的数要尽可能的小。那么相乘要最小肯定是一位数乘一位数。除的话有两种情况，可能除一位数最小可能除两位数最小，因为一位数乘一位数不可能得到三位数，如果除一个三位数会浪费了一位，然后就是加的部分了，加得到的数最大肯定是一位数加多位数或者多位数加一位数，那么现在答案很明显就只有4种情况，4种求出来取最大就是答案了，下面给代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<queue>
#include<utility>
#include<map>
#define maxn 25
typedef long long LL;
using namespace std;
const int mod=1e9+7;;
char s[maxn];
int a[maxn];
LL getnum(int l,int r){
    LL ans=0;
    for(int i=l;i<=r;i++){
        ans=(ans*10)+a[i];
    }
    return ans;
}
int main(){
    int t;
    scanf("%d",&t);
    for(int tcase=1;tcase<=t;tcase++){
        scanf("%s",s);
        int len=strlen(s);
        for(int i=0;i<len;i++){
            a[i]=s[i]-'0';
        }
        LL ans=a[0]+getnum(1,len-4)-a[len-3]*a[len-2]/a[len-1];
        ans=max(ans,getnum(0,len-5)+a[len-4]-a[len-3]*a[len-2]/a[len-1]);
        if(len>5){
            int div=getnum(len-2,len-1);
            ans=max(ans,a[0]+getnum(1,len-5)-a[len-4]*a[len-3]/div);
            ans=max(ans,getnum(0,len-6)+a[len-5]-a[len-4]*a[len-3]/div);
        }
        printf("Case #%d: %lld\n",tcase,ans);
    }
}