1023 Have Fun with Numbers

古城微笑少年丶 2024-04-08 09:53 70阅读 0赞

## 1023 Have Fun with Numbers ##

### 0、题目 ###

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

### 1、大致题意 ###

给出一个长度不超过20的整数，问这个整数两倍后的数位是否为原数位的一个排列。不管是yes还是no最后都要输出整数乘以2的结果

### 2、基本思路 ###

简单题

### 3、AC代码 ###

#include <cstdio>
    #include <string.h>
    using namespace std;
    int book[10];
    int main() {
        
    	char num[22];
    	scanf("%s", num);
    	int flag = 0, len = strlen(num);
    	for(int i = len - 1; i >= 0; i--) {
        
    		int temp = num[i] - '0';
    		book[temp]++;
    		temp = temp * 2 + flag;
    		flag = 0;
    		if(temp >= 10) {
        
    			temp = temp - 10;
    			flag = 1;
    		}
    		num[i] = (temp + '0');
    		book[temp]--;
    	}
    	int flag1 = 0;
    	for(int i = 0; i < 10; i++) {
        
    		if(book[i] != 0)
    			flag1 = 1;
    	}
    	printf("%s", (flag == 1 || flag1 == 1) ? "No\n" : "Yes\n");
    	if(flag == 1) printf("1");
    	printf("%s", num);
    	return 0;
    }