一道acm题目 报错是wrong answer 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 significan

一道acm题目 报错是wrong answer 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 significan
一道acm题目 报错是wrong answer
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.
Each line of input contains two unsigned integers less than 10 digits.The last line of input contains 0 0.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
Output for Sample Input
No carry operation.
3 carry operations.
1 carry operation.
作答
#include
#include
using namespace std;
int main()
{
\x09string x,y;
\x09while(getline(cin,x,' '),getline(cin,y)&&x!="0"||y!="0")
\x09{
\x09
\x09\x09int n=x.length();
\x09\x09int m=y.length();
\x09\x09int max=(n>=m)?n:m;
\x09\x09int min=(n=m)?x:y;
\x09\x09string mins=(n0;i--)
\x09\x09{
\x09\x09\x09if(maxs[max-1]+mins[min-1]>=106||maxs[max-1]>=58)
\x09\x09\x09{
\x09\x09\x09\x09count++;
\x09\x09\x09\x09maxs[max-2]++;
\x09\x09\x09\x09max--;
\x09\x09\x09\x09min--;
\x09\x09\x09
\x09\x09\x09}
\x09\x09\x09else
\x09\x09\x09{
\x09\x09
\x09\x09\x09\x09max--;
\x09\x09\x09\x09min--;
\x09\x09\x09}
\x09\x09
\x09\x09}
\x09 if(count>1)
\x09 {
\x09 \x09cout

一道acm题目 报错是wrong answer 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 significan
最明显的错误,就是
No carry operation.


这些最后面都有一个句号.你的输出没有.


这题是算有多少次进位吧,你是模拟加法运算,
if(maxs[max-1]+mins[min-1]>=106||maxs[max-1]>=58) 这些别人只能猜出来是什么意思,


maxs[max-2]++;这个百分百溢出 .当max = 1的时候  ,即原来的长度是1
没必要用字符串.
#include <iostream>
using namespace std;

int main()
{
\x09unsigned ua, ub;
\x09while(cin >> ua >> ub)
\x09{
\x09\x09if( ua == 0 && ub == 0) break;
\x09\x09// 只要依次抽出最后一位,看下有没有进位就可以.
\x09\x09int ntimes = 0;
\x09\x09int nCarry = 0;
\x09\x09// 循环的条件 ua,ub有一个不为0 
\x09\x09while( ua != 0 || ub != 0)
\x09\x09{
\x09\x09\x09unsigned a = ua % 10;
\x09\x09\x09unsigned b = ub % 10;
\x09\x09\x09if( a + b + nCarry >= 10)
\x09\x09\x09{
\x09\x09\x09\x09nCarry = 1;
\x09\x09\x09\x09ntimes ++ ;
\x09\x09\x09}
\x09\x09\x09else
\x09\x09\x09\x09nCarry = 0;
\x09\x09\x09if(ua == 0 && nCarry == 0) break;
\x09\x09\x09if(ub == 0 && nCarry == 0) break;\x09// 如果其中有一个为0,且没有进位,则退出
\x09\x09\x09ua /= 10;
\x09\x09\x09ub /= 10;
\x09\x09}
\x09\x09if( nCarry == 1)
\x09\x09\x09ntimes ++;
\x09\x09if(ntimes == 0)
\x09\x09\x09cout << "No carry operation." << endl;
\x09\x09else if(ntimes == 1)
\x09\x09\x09cout << "1 carry operation." << endl;
\x09\x09else
\x09\x09\x09cout << ntimes << " carry operation." << endl;

\x09}
\x09return  0;
}