Problem G: If We Were a Child Again

最新推荐文章于 2022-10-19 21:09:47 发布

转载最新推荐文章于 2022-10-19 21:09:47 发布 · 134 阅读

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原文链接：http://www.cnblogs.com/NYNU-ACM/p/4236900.html

本文介绍了一个大学算术项目的解决方案，重点在于实现基本的整数除法和取模操作。项目背景是一名学生需要制作一个计算器，但不想独立完成，于是求助于朋友编写相关代码。文章提供了详细的C++代码示例，展示了如何处理任意长度的非负整数输入，并执行除法和取模运算。

Problem G: If We Were a Child Again
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 18 Solved: 14
[Submit][Status][Web Board]
Description
The Problem

The first project for the poor student was to make a calculator that can just perform the basic arithmetic operations.

But like many other university students he doesn’t like to do any project by himself. He just wants to collect programs from here and there. As you are a friend of him, he asks you to write the program. But, you are also intelligent enough to tackle this kind of people. You agreed to write only the (integer) division and mod (% in C/C++) operations for him.

Input
Input is a sequence of lines. Each line will contain an input number. One or more spaces. A sign (division or mod). Again spaces. And another input number. Both the input numbers are non-negative integer. The first one may be arbitrarily long. The second number n will be in the range (0 < n < 231).

Output
A line for each input, each containing an integer. See the sample input and output. Output should not contain any extra space.

Sample Input
110 / 100
99 % 10
2147483647 / 2147483647
2147483646 % 2147483647
Sample Output
1
9
1
2147483646

my answer:

#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int main()
{
    char a[100];
    int t[101];
    int sum,b,i;
    char c;
    while(cin>>a)
    {
        cin>>c;
        cin>>b;
        int t1=strlen(a);
        sum=a[0]-'0';
 
        for( i=0;i!=t1;i++){
            t[i]=sum/b;
            sum=sum%b;
            if(sum<b&&i<t1-1){
            sum=(sum*10+a[i+1]-'0');
 
            }
        }
        if(c=='%')
            cout<<sum<<endl;
        else{
        int start=0;
        while(!t[start])start++;
        if(start==i)cout<<1<<endl;
        else {
        for(int j=start;j<i;j++)
            cout<<t[j];
        cout<<endl;
        }
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/NYNU-ACM/p/4236900.html