编写题解 1201: 回文数(一)

解题思路:数字转换为字符串 倒置和判断回文两个函数 

注意事项:

参考代码:

#include "stdio.h"

#include "string.h"

#include "stdlib.h"

void NIZHI(char NUM[100]) //倒置

{

char a[100];

int h = strlen(NUM);

int j = 0;

for (int i = h - 1; i >= 0; i--) 

{

a[j] = NUM[i]; j++;

}

a[j] = '\0';

strcpy(NUM, a);

}

int HUIWEN(char NUM[100])//回文

{

int h = strlen(NUM);

for (int i = 0; NUM[i] != '\0'; i++)

{

if (NUM[i] == NUM[h - 1]) { h--; continue; }

if (NUM[i] != NUM[h - 1]) { break; }

}

if (h == 0) { return 1; }

if (h != 0) { return 0; }

}

int main() 

{

int p; scanf("%d", &p); 

for (int j = 0; j < p; j++)

{

int n;

scanf("%d", &n);

int k = 0;

while (1)

{

char NUM[100];

int h;

h = sprintf(NUM, "%d", n);//数字转换为字符

if (HUIWEN(NUM) == 1)

{

printf("%d\n", k); break;

}

if (HUIWEN(NUM) == 0)

{

int x;

x = atoi(NUM);

NIZHI(NUM);

x += atoi(NUM);//变回数字

n = x;

k++;

}

if (k > 8) { printf("0\n"); break; }

}

}

}