Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10724		Accepted: 3980

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 456555 555123 5940 0

Sample Output

No carry operation.3 carry operations.1 carry operation. 大坑 CODE:

#include 
     
      #include 
      
       #include 
       
        #define REP(i, s, n) for(int i = s; i <= n; i ++)#define REP_(i, s, n) for(int i = n; i >= s; i --)#define MAX_N 10 + 5using namespace std;char a[MAX_N], b[MAX_N];int int_a[MAX_N], int_b[MAX_N], la, lb;int main(){    while(scanf("%s%s", a + 1, b + 1) != EOF){        if(a[1] == '0' && b[1] == '0') break;        memset(int_a, 0, sizeof(int_a)); memset(int_b, 0, sizeof(int_b));        la = strlen(a + 1), lb = strlen(b + 1);        REP(i, 1, la) int_a[la - i + 1] = a[i] - '0';        REP(i, 1, lb) int_b[lb - i + 1] = b[i] - '0';                    int i = 1, x = 0, res = 0;        while(i <= la || i <= lb){            x = int_a[i] + int_b[i] + x;            x /= 10;            if(x != 0) res ++;            i ++;        }                if(res == 0) printf("No carry operation.\n");        else if(res == 1) printf("%d carry operation.\n", res);        else printf("%d carry operations.\n", res);    }    return 0;}