PAT_甲级_1023

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1023 Have Fun with Numbers

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

大整数乘法

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
char num[25];
int num_int[25];
int count_num[10];
int result[25] = {0};
int main()
{
    cin >> num;
    memset(count_num, 0, sizeof(count_num));
    memset(num_int, 0, sizeof(num_int));
    int len = strlen(num);
    for (int i = 0; i < len; i++) {
        num_int[len - i - 1] = num[i] - '0';
    }
    for (int i = 0; i < len; i++) {
        count_num[num_int[i]]++;
    }
    int carry = 0;
    int len_num = 0;
    for (int i = 0; i < len; i++) {
        int temp = num_int[i] * 2 + carry;
        result[len_num++] = temp % 10;
        carry = temp / 10;
    }
    while (carry != 0) {                 // 如果最后还进位
        result[len_num++] = carry % 10;
        carry = carry / 10;
    }
    for (int i = 0; i < len_num; i++) {
        count_num[result[i]]--;
    }
    int flag = 0;
    for (int i = 0; i < 10; i++) {
        if (count_num[i] != 0) {
            cout << "No" << endl;
            flag = 1;
            break;
        }
    }
    if (flag == 0) {
        cout << "Yes" << endl;
    }
    for (int i = len_num - 1; i >= 0; i--) {
        cout << result[i];
    }
    return 0;
}