EASY_ZJU_PAT_ADVANCED LEVEL_1023 大数乘法

最新推荐文章于 2020-03-04 22:47:04 发布

原创最新推荐文章于 2020-03-04 22:47:04 发布 · 744 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#C++ #OJ #ZJU #PAT

ZJU_PAT 专栏收录该内容

29 篇文章

订阅专栏

本文介绍了一个编程问题：检查一个给定的数字在翻倍之后，其数字是否仅由原始数字的排列组成。通过字符串转换为整数数组，并进行翻倍操作后比较两个数组是否一致来实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

32000 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

/************************************************
	@ AUTHOR	: GAOMINQUAN
	@ DATA		: 2014 - 2 - 24
	@ MAIL		: ENSOLEILLY@GMAI.COM
	@ HARD		: EASY **


/************************************************/




#include<iostream>
#include<vector>
#include<string>
#include<algorithm>


using namespace std;




vector<int> reverse_vector(vector<int> perVec){
	int length = perVec.size() - 1;


	for(int i = 0; i<=length/2; i++){  //NOTICE IT'S LIKE THE PRIMER CACULATE, < AND <=
		swap(perVec[i],perVec[length-i]);
	}
	return perVec;
}


vector<int> change_to_bits(string inputNum){
	vector<int> bits;
	for(int stringI = 0; stringI < inputNum.size(); stringI++){
		int tempBits = inputNum[stringI] - '0';
		bits.push_back(tempBits);
	}


	return bits;
}


vector<int> mutiply(vector<int> numbers){
	bool addBit = false;
	vector<int> doubleNum;
	for(int i = numbers.size() - 1; i>=0; i--){
		int tempDouble = numbers[i] * 2 + addBit;
		if(tempDouble>9){
			tempDouble -= 10;
			addBit = true;
		}else{
			addBit = false;
		}
		doubleNum.push_back(tempDouble);
	}
	if(addBit){
		doubleNum.push_back(addBit);  //NOTICE HERE !!!!!
	}
	return doubleNum;
}


bool checkSame(vector<int> v1,vector<int> v2){
	sort(v1.begin(),v1.end());
	sort(v2.begin(),v2.end());


	bool same = (v1.size() == v2.size());


	if(v1.size() == v2.size()){
		for(int i = 0; i<v1.size(); i++){
			if(v1[i] != v2[i]){
				same = false;
				break;
			}
		}
	}


	return same;
}




int main(){
	string input = "123456789";
	
	cin>>input;


	string outputs[] = {"No","Yes"};


	vector<int> bits = change_to_bits(input);


//	vector<int> doubleBits = mutiply(bits);
	vector<int> doubleBits = reverse_vector(mutiply(bits));
	cout<<outputs[checkSame(bits,doubleBits)]<<endl;


	for(int i = 0; i<doubleBits.size(); i++){
		cout<<doubleBits[i];
	}cout<<endl;


	
	return 0;
}