POJ 2562 Primary Arithmetic 進位統計

原題 http://poj.org/problem?id=2562

題目：

Primary Arithmetic

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 11008 Accepted: 4048

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.

思路：

求兩個數相加進位了多少次。

此處我們可以效仿高精度的做法，進位的同時ans++。

注意輸出格式，進1位和多位的差別，有個s。

代碼：

#include <iostream>
#include"string.h"
#include"cstdio"
#include"stdlib.h"
#include"algorithm"
using namespace std;
typedef long long int lint;


int main()
{
    const int N=;
    lint a,b;
    while(cin>>a>>b)
    {
        if(a==&&b==)  break;
        lint s1[N];
        lint s2[N];
        memset(s1,,sizeof(s1));
        memset(s2,,sizeof(s2));
        int l1=;
        while(a>)
        {
            s1[l1]=a%;
            l1++;
            a=a/;
        }
        int l2=;
        while(b>)
        {
            s2[l2]=b%;
            l2++;
            b=b/;
        }
        int ans=;
        for(int i=; i<max(l1,l2)+; i++)
        {
            s1[i]=s1[i]+s2[i];
            if(s1[i]>=)
            {
                s1[i]=s1[i]%;
                s1[i+]++;
                ans++;
            }
        }
        if(ans==)  printf("No carry operation.\n");
        else if(ans==) printf("1 carry operation.\n");
        else
        printf("%d carry operations.\n",ans);

    }
    return ;
}