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  • 1350. Primary Arithmetic

    Children are taught to add multi-digit numbers from right-to-left one digit at a time.  Many find the "carry" operation - in which a 1 is carried from one digit position to be added to the next - to be a significant challenge.  Your job is to count the number of  carry operations for each of a set of addition problems so that educators may assess their difficulty.

    Each line of input contains two unsigned integers less than 10 digits.   The last line of input contains 0 0.  For each line of input except the  last you should compute and print the number of carry operations that would result from adding the two numbers, in the format shown below.

    Sample Input

    123 456
    555 555
    123 594
    0 0
    

    Output for Sample Input

    No carry operation.
    3 carry operations.
    1 carry operation.
    
    #include<iostream>
    #include<cmath>
    using namespace std;
    int main()
    {
        unsigned int m,n,sum,i;
        //cin>>m>>n;
        while(cin>>m>>n)
        {
            if(m==0&&n==0)
              break;
            unsigned int temp;
            if(m<n)
            {
                temp=m;
                m=n;
                n=temp;
            }        
            unsigned int m1,n1;
            i=0;
            sum=0;
        while(m>0)
            {
            m1=m%10;
            n1=n%10;
            if(m1+n1+i>=10)
            {
                i=1;
                sum++;
            }
            else {
                i=0;
            }
            m=m/10;
            n=n/10;
            
        }
        while(n>0)
        {
            if(n%10+i>=10)
            {
                i=1;
                sum++;
            }
            else
            {
                i=0;
            }
            n=n/10;
        }
        if(sum==0)
        {
           cout<<"No carry operation."<<endl;
        }  
        else if(sum==1)
        {
           cout<<sum<<" carry operation."<<endl;
        }  
        else if(sum>1)
        {
           cout<<sum<<" carry operations."<<endl;
        }
            //cin>>m>>n; 
        }
            return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/wft1990/p/6057589.html
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