Primary Arithmetic
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11706 | Accepted: 4287 |
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.
Source
題解：
首先用字符串輸入，反向轉化成整型，接着對應位置相加，如果相應位相加的和大於9，則需要進位。
注意：
進位數單複數的問題，還有幾個字符型常用的函數一定要牢記！
代碼：
#include<iostream>
#include <cstring>
using namespace std;
void getNumber(char str[] , int c[])
{
int i,len = strlen(str);
for(i = 0 ;i < len ;i++)
{
c[i] = str[len - 1 - i]-'0';
}
}
int main(){
char a[11],b[11];
int n[11] = {0}, m[11] = {0}, i, len, num;
while(cin>>a>>b)
{
if(strcmp(a,"0") == 0 && strcmp(b,"0")== 0)
break;
num = 0;
getNumber(a,n);
getNumber(b,m);
for(i = 0;i < 11;i++)
{
if(n[i] + m[i] > 9)
{
n[i+1] += (n[i] + m[i]) / 10;
n[i] = m[i] = 0;
num ++;
}
}
if(num == 0)
cout << "No carry operation." << endl;
if(num == 1)
cout << num << " carry operation." << endl;
if(num > 1)
cout << num << " carry operations." << endl;
}
return 0;
}