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.
Each input contains one test case. Each case contains one positive integer with no more than 20 digits.
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.
1234567899
Yes
2469135798
题目解析
本题给出一个长度再20以内的数字（由1~9组成），要求判断这个数字加倍后的新数字是不是这个数字的某一种排列，如果是的化输出Yes否则输出No，之后输出加倍后的数字。
由于数字最大位数为20位，超过了long long int的记录范围我们用数组num记录这个数字，用数组cnt记录num中1~9出现的次数，将num加倍后判断其中1~9出现的次数是否发送改变，若没有发送改变则证明加倍后的数字是原数字的某一种排列，反之则不是。
AC代码
1 #include <bits/stdc++.h> 2 using namespace std; 3 int num[22]; 4 int cnt[10]; 5 string str; 6 void toInt(){ 7 for(int i = 0; i < str.size(); i++) 8 num[i] = str[i] - ‘0‘; 9 } 10 void getCnt(){ 11 for(int i = 0; i < str.size(); i++) 12 cnt[num[i]]++; 13 } 14 bool judge(int carry){ 15 if(carry != 0) //如果最高位进位不为零，则证明加倍后的数字比原数字多一位，那么其肯定不是原数字的一个排列 16 return false; 17 for(int i = 0; i < str.size(); i++) 18 cnt[num[i]]--; 19 for(int i = 1; i <= 9; i++){ //判断新的num中1~9的数量是否和加倍前一样 20 if(cnt[i] != 0) 21 return false; 22 } 23 return true; 24 } 25 int doubleNumber(){ //将数组num加倍并返回最高位进位 26 int carry = 0; 27 for(int i = str.size() - 1; i >= 0; i--){ 28 int temp = num[i]; 29 num[i] = (2 * temp + carry) % 10; 30 carry = 2 * temp / 10; 31 } 32 return carry; 33 } 34 int main() 35 { 36 cin >> str; //输入数字 37 toInt(); //将输入的数字转化为数组 38 getCnt(); //获取数组中1~9出现的次数 39 int carry = doubleNumber(); //将num加倍carry记录最高位的进位 40 if(judge(carry)){ //判断加倍后的数字是否为原数字的某一个排列 41 printf("Yes\n"); 42 43 }else 44 printf("No\n"); 45 if(carry != 0) //判断是否需要输出进位 46 printf("%d", carry); 47 for(int i = 0; i < str.size(); i++) //输出加倍后的数组num 48 printf("%d", num[i]); 49 printf("\n"); 50 return 0; 51 }
PTA (Advanced Level)1023 Have Fun with Numbers
原文：https://www.cnblogs.com/suvvm/p/10311268.html