HDU 5938 Four Operations (水题乱搞)

最新推荐文章于 2020-11-06 21:27:10 发布

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最新推荐文章于 2020-11-06 21:27:10 发布

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本文介绍了一道编程题，要求使用四则运算使给定数字串表达式的计算结果达到最大值。通过固定符号位置的方法，对不同长度的输入字符串进行处理，并提供了完整的C++代码实现。

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Four Operations

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 413 Accepted Submission(s): 149

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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要把+-*/按顺序填入给定字符串，使式子结果最大。这个思考起来还是比较容易的，符号填入的位置都是固定的几种组合，代码里已经很清楚了。。

#include "cstring"
#include "cstdio"
#include "iostream"
#include "string.h"
#include "cmath"
#include "algorithm"
using namespace std;
char s[50];
long long getnum(int start,int end)
{
    long long sum=0;
    long long temp=1;
    for(int i=end;i>=start;i--)
    {
        sum+=temp*(s[i]-'0');
        temp*=10;
    }
    return sum;
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int k=1;k<=t;k++)
    {
        scanf("%s",s);
        int len=strlen(s);
        if(len==5)
        {
            printf("Case #%d: %d\n",k,getnum(0,0)+getnum(1,1)-getnum(2,2)*getnum(3,3)/getnum(4,4));
            continue;
        }
        long long back1=getnum(len-3,len-3)*getnum(len-2,len-2)/getnum(len-1,len-1);
        long long back2=getnum(len-4,len-4)*getnum(len-3,len-3)/getnum(len-2,len-1);
        long long front11=getnum(0,0)+getnum(1,len-4);
        long long front12=getnum(0,len-5)+getnum(len-4,len-4);
        long long front21=getnum(0,0)+getnum(1,len-5);
        long long front22=getnum(0,len-6)+getnum(len-5,len-5);
        long long ans=-1;
        ans=max(ans,front11-back1);
        ans=max(ans,front12-back1);
        ans=max(ans,front21-back2);
        ans=max(ans,front22-back2);
        printf("Case #%d: %lld\n",k,ans);

    }
}