求C的大佬，输出都是对的，错三个测试点

P1307 [NOIP2011 普及组] 数字反转

Dadao153 @ 2024-12-22 20:23:41

题目是数字反转，自己把可能的特殊值和大数小数，上限都试过了，都是对的，错了三个测试点，代码如下。

#include<stdio.h>
int i;
int s[10000];
void daoxu(long long n) {   
    for (i=0; n != 0; i++) {
        s[i] = n % 10;
        n = n / 10;
    }
}

int main() {

    long long m = 0,k=0;
    scanf("%lld", &m);

    k = m;
    if (m < 0)m = -m;
    daoxu(m);
    if (m == 0)printf("0");
    if (k >= 0) {
        for (int p = 0; p < i; p++) {
            if (s[p] != '\0')printf("%d", s[p]);
        }
    }
    else if (k < 0) {
        printf("-");
        for (int p = 0; p < i; p++) {
            if (s[p] != '\0')printf("%d", s[p]);
        }
    }

    return 0;
}

by pika_ @ 2024-12-22 20:30:45

hack:
input:
-10000000001
your:
-11
ans:
-10000000001

by hahadamntards @ 2024-12-23 22:50:40

#define _CRT_SECURE_NO_WARNINGS 1
#include <math.h>
#include <stdio.h>
int main()
{
    int n,x,a,i=0, sum = 0,b,c;
    scanf("%d", &n);
    /*求+-号*/
    x = n > 0 ? 1 : -1;
    n = fabs(n);
    c = n;
    while (c!= 0)
    {
        c /= 10;
        i++;
    }
    for (int j = i - 1; j >= 0; j--)
    {
        a = n % 10;
        n /= 10;
        sum +=a*pow(10, j);
    }
    printf("%d\n",sum*x);
    return 0;
}