1 题目
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
题目链接:
2 解题思路
给定一个数,乘以2之后,各个数位的出现次数,是否与乘之前是否相同,相同输出Yes,否认输出No,然后下一行打印出乘2之后的各个数位。
这个题目是属于超大数运算的类型,题目给出的最高数位是20位,那么即便是用最高的unsigned long long
也会面临溢出的情况,所以输入和输出,只能用string
,诸位乘2,然后再记录每一位出现的次数,相比较就行。
3 AC代码
/*
** @Brief:No.1023 of PAT advanced level.
** @Author:Jason.Lee
** @Date:2018-8-04
*/
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int digit[10];
string input;
string output;
bool doubleNumber(string input){
int flag = 0;
int temp = 0;
output.resize(input.length());
for(int i = input.length()-1;i>=0;i--){
temp = (int)(input[i]-'0');
digit[temp]++;
temp=2*temp+flag;
if(temp>9){
temp%=10;
flag = 1;
}else{
flag = 0;
}
output[i] = (temp+'0');
//cout<<"ouput :"<<output[i]<<endl;
}
if(flag){
output= "1"+output;
}
//cout<<"output : "<<output<<endl;
//cout<<"size : "<<output.length()<<endl;
for(int i = 0;i<output.length();i++){
digit[(int)(output[i]-'0')]--;
}
for(int i=0;i<10;i++){
if(digit[i]!=0){
return false;
}
}
return true;
}
int main(){
cin>>input;
cout<<(doubleNumber(input)? "Yes":"No")<<endl;
cout<<output<<endl;
return 0;
}