ZOJ Problem Set–1874 Primary Arithmetic

Time Limit: 2 Seconds      Memory Limit: 65536 KB


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.


Source: University of Waterloo Local Contest 2000.09.23

  1: #include<iostream>
  2: #include<string>
  3: #include<algorithm>
  4: using namespace std;
  5: int countCarries(const string& pre1, const string& pre2)
  6: {
  7:   size_t loop = pre1.length() > pre2.length() ? pre1.length():pre2.length();
  8:   bool carry = false;
  9:   int carryCount = 0;
 10:   short tempResult;
 11:   size_t i = 0;
 12:   for(i = 0; i < loop; i++)
 13:   {
 14:     tempResult = (pre1[i] - '0')*(i < pre1.length()) + (pre2[i] - '0')*(i < pre2.length()) + (short)carry;
 15:     carry = (tempResult >= 10);
 16:     carryCount += carry;
 17:   }
 18:   return carryCount;
 19: }
 20: int main(void)
 21: {
 22:   string op1, op2;
 23:   while(cin>>op1>>op2 && (op1 != "0" || op2 != "0"))
 24:   {
 25:     reverse(op1.begin(), op1.end());
 26:     reverse(op2.begin(), op2.end());
 27:     int c = countCarries(op1, op2);
 28:     if(c == 0)
 29:     {
 30:       cout<<"No carry operation."<<endl;
 31:     }
 32:     else if( c == 1)
 33:     {
 34:       cout<<c<<" carry operation."<<endl;
 35:     }
 36:     else
 37:     {
 38:       cout<<c<<" carry operations."<<endl;
 39:     }
 40:   }
 41:   return 0;
 42: }
posted @ 2012-05-10 13:50  Gavin Lipeng Ma  阅读(499)  评论(0编辑  收藏  举报