【PAT】A1023 Have Fun with Numbers (20point(s))

Author: CHEN, Yue
Organization: 浙江大学
Time Limit: 400 ms
Memory Limit: 64 MB
Code Size Limit: 16 KB

A1023 Have Fun with Numbers (20point(s))

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

Code

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <vector>
using namespace std;
char num[21];
vector<int> vi;
int htable[10], flag = 1;
int main(){
	memset(htable, 0, sizeof(htable));		// 表初始置零
	scanf("%s", num);
	for (int i = 0; i < strlen(num); i++){
		htable[num[i] - '0']++;				// 记录每一个数字出现的次数
		vi.push_back((num[i] - '0') * 2);
	}
	for (int i = vi.size()-1; i > 0; i--){	// 这里我们不对vi[0]进位，因为如果vi[0]需要进位，那么结果一定为“No”
		if (vi[i] / 10){					// 由于本题只是将每位数据乘二，所以在一个vi[i]中不会出现三位数
			vi[i - 1] += vi[i] / 10;
			vi[i] = vi[i] % 10;
		}
		htable[vi[i]]--;					// 数字出现过就把对应表的值--
	}
	if (vi[0] / 10){						// 如果最高位需要进位，即乘二后数字加长一位
		printf("No\n");
		for (int i = 0; i < vi.size(); i++)
			printf("%d", vi[i]);
		printf("\n");
		return 0;
	}
	htable[vi[0]]--;
	for (int i = 0; i < vi.size(); i++){
		if (htable[vi[i]]){					// 有数字多出现了，当然这种情况下表中必有负值
			printf("No\n");
			flag = 0;
			break;
		}
	}
	if (flag)	printf("Yes\n");
	for (int i = 0; i < vi.size(); i++)	printf("%d", vi[i]);
	printf("\n");
	return 0;
}

Analysis

-已知一个k位的数字。

-求，将这个数字乘以2后，如果新的数字里的每一位和原数字的每一位中各种数字（0-9）出现的次数完全相同，则输出Yes，否则输出No。