pat00-自测4. Have Fun with Numbers (20)

00-自测4. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 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

---

提交代码

 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <queue>
 #include <cmath>
 #include <iostream>
 using namespace std;
 int ti[],num1[],num2[];
 char num[];
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     scanf("%s",num);

     //cout<<num<<endl;

     int i;
     for(i=;i<strlen(num);i++){
         num1[i]=num[i]-'';
         ti[num1[i]]++;
         num2[i]=*num1[i];
     }
     int k=,t;
     for(i=strlen(num)-;i>=;i--){
         t=num2[i]+k;
         num2[i]=t%;
         ti[num2[i]]--;
         k=t/;
     }
     bool can=false;
     if(!k){
         for(i=;i<=;i++){
             if(ti[i]){
                 break;
             }
         }
         if(i==){
             can=true;
         }
     }
     if(can){
         printf("Yes\n");
     }
     else{
         printf("No\n");
     }
     if(k){
         printf("%d",k);
     }
     for(i=;i<strlen(num);i++){
         printf("%d",num2[i]);
     }
     printf("\n");
     return ;
 }