1023 Have Fun with Numbers (20)（20 分）
题目描述：
Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!
Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.
输入格式：
Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.输出格式：
For each test case, first print in a line “Yes” if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or “No” if not. Then in the next line, print the doubled number.
题目大意：
这道题目主要是判断是否一个数字在乘以2之后，他的结果是否仍然为原来数字的组合，比如说
22*2=44，显然，原来的数字是2个2的组合，而结果是2个4的组合，不一样，因此输出No。
解题方法：
这道题目在解题上没有难度，直接把输入的数字用char数组存储，然后从数组的末尾开始，做乘法，不过需要考虑进位的问题，具体可以看我代码。
易错点：
1. 不要漏了最后一个进位，如999只用for循环会漏掉1998中的1
程序：
#include <stdio.h>
#include <string.h>
int main(int argc, char const *argv[])
{
char num[21];
int douNum[22], check[10] = {0};
int douIdx = 0, res, add = 0, flag = 0;
scanf("%s",num); /* 输入测试样例 */
for (int i = strlen(num) - 1; i >= 0; i--)
{
res = (num[i]-'0') * 2 + add;
if (res >= 10)
{
res %= 10;
add = 1; /* 进位置1 */
}
else
add = 0; /* 进位置0 */
douNum[douIdx++] = res;
}
if (add == 1) /* 如果有36进位比如(9->18) */
douNum[douIdx++] = 1;
for (int i = 0; i < douIdx; i++)
check[douNum[i]]++; /* 对应数字出现次数+1 */
for (int i = 0; i < strlen(num); i++)
check[num[i]-'0']--; /* 对应数字剩余出现次数-1 */
for (int i = 0; i < 10; i++)
if (check[i] != 0) /* 如果出现该数字出现次数不相同 */
{
flag = 1;
break;
}
if (flag == 1)
printf("No\n");
else
printf("Yes\n");
for (int i = douIdx-1; i >= 0; i--)
printf("%d", douNum[i]);
return 0;
}如果对您有帮助，帮忙点个小拇指呗~