A1023 Have Fun with Numbers（20分）PAT 甲级(Advanced Level) Practice（C++）满分题解【大整数相加】

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

题意分析：

整数乘以二转换成两个相同的大整数相加，然后使用两个map来存放两个字符串中的0-9的每个数字的个数，最后将两个map进行逐数字比较

代码如下：

#include<bits/stdc++.h>
using namespace std;
string multiple2(string s)
{
	reverse(s.begin(), s.end());
	string s2 = s, res;
	int carry = 0;
	for (int i = 0; i < s2.length()|| i < s.length(); i++) {
		int temp = (s2[i]-'0') + (s[i]-'0') + carry;//对应位相加，再加上进位
		res += temp % 10 + '0';//相加个位数为当前位的结果
		carry = temp / 10;
	}
	if (carry != 0)res += carry + '0';
	reverse(res.begin(), res.end());
	return res;
}
int main()
{
	string s,s2;
	map<char, int> count,count2;
	cin >> s;
	s2 = multiple2(s);
	for (int i = 0; i < s.length(); i++) {
		count[s[i]]++;
	}
	for (int i = 0; i < s2.length(); i++) {
		count2[s2[i]]++;
	}
	int i;
	for (i = 0; i <= 9; i++) {
		if (count[i+'0'] != count2[i+'0'])
			break;
	}
	if (i > 9)cout << "Yes" << endl;
	else cout << "No" << endl;
	cout << s2 << endl;
	return 0;
}

运行结果如下：