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  • Problem G: If We Were a Child Again

    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;
    }
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  • 原文地址:https://www.cnblogs.com/NYNU-ACM/p/4236900.html
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