PAT

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.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

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.

Sample Input:

1234567899

Sample Output:

Yes2469135798

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给定条件：
1.给定一个不超过20位的数字，只有1-9十个数字

要求：
1.将这个数字乘以2
2.得到的新的数字是否和原来的给定的数包含的数字都是相同的，并且每个数字出现的次数也都是相同的

求解：
1.记录每个数字出现的次数
2.用数组模拟乘法运算的到结果
3.遍历得到的新数字，进行判断

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#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n;int num[25], book[25];char input[25];int main() {fill(book, book+25, 0);scanf("%s", input);n = strlen(input)-1;for(int i = n; i >= 0; i--) {num[n-i] = input[i] - '0';book[num[n-i]]++;}int up = 0;for(int i = 0; i <= n; i++) {int sum = 2 * num[i] + up;num[i] = sum % 10;up = sum / 10;}if(up != 0){num[++n] = up;}int flag = 1;for(int i = 0; i <= n; i++) {book[num[i]]--;if(book[num[i]] < 0){printf("No\n");flag = 0;break;}}if(flag) printf("Yes\n");for(int i = n; i >= 0 ; i--) {printf("%d", num[i]);}printf("\n");return 0;}

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