Contest 3:Problem A: Number and its reverse

最新推荐文章于 2018-12-20 15:34:54 发布

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此题注意c中无reverse函数，用模的方法进行reverse操作。计算反转数时忘注意回文数，导致了最开始想减小计算域，结果从1开始也不会超时。C'est tout.

Description

Give you a number M (9999<=M<=9999999).

Output the number of total numbers such that the sum of this number and its reverse is M.

Input

There are several lines. And each line contains a positive number M (9999<=M<=9999999).

Output

Output the number of total numbers such that the sum of this number and its reverse is M.

Output 0 if there is no such numbers.

For example,

Input

1333

2444

Output

Because 419, 518, 617, 716, 815, 914 are total numbers such that the sum of this number and its reverse is 1333. (419+914=1333)

Sample Input

1333
2444

Sample Output

6
0

HINT

#include <stdio.h>
#include <stdlib.h>
int reverse(int x)
{
    int r = 0;
    while(x)
    {
        r=r*10+x%10;
        x/=10;
    }
    return r;
}
int main()
{   int n,j,i,num;
     while (scanf("%d",&n)!=EOF){
    num=0;
    for(i=1;i<=n;i++){
        j=reverse(i);
    if(n==j+i)
        num+=1;
    }
    printf("%d\n",num);}
    return 0;
}