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
题解：
首先用字符串输入，反向转化成整型，接着对应位置相加，如果相应位相加的和大于9，则需要进位。
注意：
进位数单复数的问题，还有几个字符型常用的函数一定要牢记！
代码：
#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;
}