PAT 甲级 1023 Have Fun with Numbers（20 分） C++版

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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 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:

Yes
2469135798

判断一个k位数的两倍是不是原来那个数各个位数的全排列

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#include <cstdio>
#include <cstring>

int main() {
    char s[22];
    scanf("%s", s);
    int *pers = new int[10], temp = 0, *pree = new int[10];
    for (int i = strlen(s) - 1; i >= 0; i--) {
        pers[s[i] - '0']++;
        temp += 2 * (s[i] - '0');
        s[i] = temp % 10 + '0';
        pree[s[i] - '0']++;
        temp /= 10;
    }
    if (temp != 0) pree[temp + '0']++;
    bool state = true;
    for (int i = 1; i <= 9; i++) {
        if (pers[i] != pree[i]) {
            state = false;
        }
    }
    printf("%s\n", state ? "Yes" : "No");

    if (temp != 0) printf("%d", temp);
    printf("%s", s);
    delete[] pree;
    delete[] pers;
    return 0;
}