PTA (Advanced Level) 1023 Have Fun with Numbers

Have Fun with Numbers

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

 #include <bits/stdc++.h>
 using namespace std;
 int num[];
 int cnt[];
 string str;
 void toInt(){
     for(int i = ; i < str.size(); i++)
         num[i] = str[i] - '';
 }
 void getCnt(){
     for(int i = ; i < str.size(); i++)
         cnt[num[i]]++;
 }
 bool judge(int carry){
     if(carry != )  //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
         return false;
     for(int i = ; i < str.size(); i++)
         cnt[num[i]]--;
     for(int i = ; i <= ; i++){    //判断新的num中1~9的数量是否和加倍前一样
         if(cnt[i] != )
             return false;
     }
     return true;
 }
 int doubleNumber(){ //将数组num加倍并返回最高位进位
     int carry = ;
     for(int i = str.size() - ; i >= ; i--){
         int temp = num[i];
         num[i] = ( * temp + carry) % ;
         carry =  * temp / ;
     }
     return carry;
 }
 int main()
 {
     cin >> str; //输入数字
     toInt();    //将输入的数字转化为数组
     getCnt();   //获取数组中1~9出现的次数
     int carry = doubleNumber(); //将num加倍carry记录最高位的进位
     if(judge(carry)){   //判断加倍后的数字是否为原数字的某一个排列
         printf("Yes\n");

     }else
         printf("No\n");
     if(carry != )  //判断是否需要输出进位
         printf("%d", carry);
     for(int i = ; i < str.size(); i++) //输出加倍后的数组num
         printf("%d", num[i]);
     printf("\n");
     return ;
 }