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

题目大意

题目给定一个大整数，将其乘二，并检验其是否是原数数字的不同排列。

解题思路

1. 用字符串形式输入大整数，并将各位数字计数；
2. 大整数乘法将其乘二；
3. 检验该结果是否为原数字不同排序，输出结果并返回0值；

代码

#include<stdio.h>
void check(char a[],char b[]){
    int i=0,num[10]={0,0,0,0,0,0,0,0,0,0};
    while(a[i]){
        num[a[i++]-'0']++;
    }
    if(b[i]){
        printf("No\n");
        return;
    }
    while(i){
        num[b[--i]-'0']--;
        if(num[b[i]-'0']<0){
            printf("No\n");
            return;
        }
    }
    printf("Yes\n");
    return;
}

int main(){
    char n1[30],n2[30];
    int len=0,sum,i,k=0,c=0;
    scanf("%s",&n1);
    while(n1[len]){
        len++;
    }
    for(i=len-1;i>=0;i--){
        sum=(n1[i]-'0')*2+c;
        c=sum/10;
        sum=sum%10;
        n2[k++]=sum+'0';
    }
    if(c==1){
        n2[k++]='1';
    }
    n2[k]=0;
    check(n1,n2);
    while(k){
        printf("%c",n2[--k]);
    }
    printf("\n");
    return 0;
}

运行结果