1023. Have Fun with Numbers (20)

1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

Yes
2469135798



PS:大数模拟乘法即可，但是有一个坑就是输入有前置0，即001234567899，输出不能有前置0

#include<bits/stdc++.h>
using namespace std;

int num_int1[10],num_int2[10];
char num[30],num2[30];

int main()
{
	int carry ;
	
	while(cin >> num){
		memset(num2,0,sizeof(num2)); 
		memset(num_int1,0,sizeof(num_int1));
		memset(num_int2,0,sizeof(num_int2)); 
		for(int i = 0; i < strlen(num); i++){
			num_int1[num[i]-'0']++;
		}
		carry = 0;
		for(int i = strlen(num)-1; i >= 0; i--){
			int n = num[i]-'0';
			int tmp = 2*n+carry;
			num2[i] = tmp%10+'0';
			num_int2[tmp%10]++;
			carry = tmp / 10;
		}	
		
		if (carry != 0){
			cout << "No" << endl;
			cout << carry  ;
		}
		else{
			int flag = 1;
			for(int i = 0; i < 10; i++){
				if (num_int1[i] != num_int2[i]){
					cout << "No" << endl;
					flag = 0;
					break;
				}
			}
			if (flag)
				cout << "Yes" << endl;
			
		}
		bool z = 0;
	    for(int i = 0;i < strlen(num);++i)
	    {
	        if(num2[i]-'0' == 0 && z == 0)
	            continue;
	        if(num2[i]-'0') z = 1;
	        printf("%d",num2[i]-'0');
	    }
		cout << endl;
	} 
	
	
	return 0;
}