poj 2562 Primary Arithmetic

Primary Arithmetic

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8356		Accepted: 3122

Description

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.

Input

Each line of input contains two unsigned integers less than 10 digits. The last line of input contains 0 0.

Output

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

Sample Output

No carry operation.
3 carry operations.
1 carry operation.

#include<iostream>
using namespace std;
int main()
{
char a[11];
char b[11];
int num1[10];
int num2[10];
int mark[10];
int sum;
int i,k;
int sizea,sizeb;
while(1)
    {
        cin>>a;
        cin>>b;
if(strcmp(a,"0")==0&&strcmp(b,"0")==0)
break;
for(k=0;k<10;k++)
        {
            mark[k]=0;
            num1[k]=0;
            num2[k]=0;
        }
        sizea=strlen(a)/sizeof(a[0]);
        sizeb=strlen(b)/sizeof(b[0]);
        k=0;
for(i=sizea-1;i>=0;i--)
        {
            num1[k]=(int)a[i]-'0';
            k++;
        }
        k=0;
for(i=sizeb-1;i>=0;i--)
        {
            num2[k]=(int)b[i]-'0';
            k++;
        }
        sum=0;
for(i=0;i<10;i++)
        {
if(i==0)
            {
if(num1[i]+num2[i]>=10)
                    mark[i]=1;
            }
else if(num1[i]+num2[i]+mark[i-1]>=10)
            {
                mark[i]=1;
            }    
        }
for(k=0;k<10;k++)
            sum+=mark[k];
if(sum==0)
            cout<<"No carry operation."<<endl;
else if(sum==1)
            cout<<sum<<" carry operation."<<endl;
else
            cout<<sum<<" carry operations."<<endl;
    }
return 0;
}

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/22/2258590.html