【PAT甲级】1023 Have Fun with Numbers 累加后组成数相同吗——模拟加法进位运算

【题意】

输入一个不超20位的数字，计算其两倍的结果，然后判断其两倍的结果与初始数的各个数位，不同数字出现次数是否相同，然后再输出初始数两倍的结果。

PS：注意初始数long long（最多19位）可能也存不下20位的情况，int（最多10位）更不用说了；

【原题链接】

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

【题解】 

思路一：

用string读入，vector来后续处理及存储；设置一个变量t来处理进位；而各数位上不同数字的计数比较，可以通过排序后直接比较容器内值是否相同来实现；需要注意遍历、计算、输出过程中的正确位序。

#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

int main(){
	string s;
	cin>>s;   //以字符串读入及存储
	
	vector<int> a;
	for(int i=s.size()-1;i>=0;i--){  //从低位到高位读入vevtor a，注意将字符转为数字(减去一个偏移量)
	    a.push_back(s[i]-'0');
	}
	
	vector<int> aa;   //用aa来存两倍结果
	int t=0,c;
	for(int i=0;i<a.size();i++){
	    c=a[i]+a[i]+t;   //t为进位数，0或1；
	    aa.push_back(c%10);  //结果当前位读入
	    t=c/10;  //存可能的进位
	}
	if(t) aa.push_back(t);  //处理最后一位可能的进位
	
	vector<int> b=aa;  //复制vector aa到vector b (重排序后要输出原来的aa)
	sort(a.begin(),a.end());  //通过重排序后vector的直接比较，可以判断各个数字的数目是否相同
	sort(aa.begin(),aa.end());
	
	if(aa==a) cout<<"Yes"<<endl;  
	else cout<<"No"<<endl;
	for(int i=b.size()-1;i>=0;i--){  //因为是将2倍的结果从低位到高位运算后直接存入，逆序输出结果，才是从高位到低位序
	    cout<<b[i];                  
	}
	
	return 0;
}