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

大整数相乘

1. 原文

原题

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

2. 解析

大数乘法运算，从len-1位开始

3. AC代码

#include<cstdio>
#include<cstring>
char str[30];
int ans[30]={};
int exist[30]={};
int main(){
	scanf("%s",str);
	int len = strlen(str);
	int idx = 0;
	int carry = 0;
	for (int i = len-1; i >= 0; i--)
	{
		int temp = (str[i]-'0')*2+carry;
		ans[idx++]=temp%10;
		carry = temp/10;
	}
	while(carry!=0){
		ans[idx++]=carry%10;
		carry/=10;
	}
	if (len!=idx)
	{
		printf("No\n");
	}else{
		for (int i = 0; i < len; ++i)
		{
			exist[str[i]-'0']++;
			exist[ans[i]]--;
		}
		int flag = 0;
		for (int i = 0; i < 10; ++i)
		{
			if (exist[i]>0)
			{
				flag = 1;
				break;
			}
		}
		if (flag == 0)
		{
			printf("Yes\n");
		}else{
			printf("No\n");
		}

	}
	for (int i = idx-1; i >= 0; i--)
	{
		printf("%d",ans[i]);
	}
	printf("\n");

	
	return 0;
}