1023. Have Fun with Numbers (20)-（大整数乘法）

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

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){    int a[25],b[25];    char str[25],str2[25];    scanf("%s",str);    int len=strlen(str);    for(int i=0;i<len;i++){        a[i]=str[len-1-i]-'0';    }    int i,carry=0;    for(i=0;i<len;i++){        b[i]=(a[i]*2+carry)%10;        carry=(a[i]*2+carry)/10;    }    if(carry){        b[i++]=carry;    }    int b_len=i;    for(int i=0;i<b_len;i++){        str2[i]=b[b_len-1-i]+'0';    }    str2[b_len]='\0';//切记不能忘记字符串结束字符     sort(str,str+len);    sort(str2,str2+b_len);//  printf("%s %s\n",str,str2);    if(strcmp(str,str2)==0){        printf("Yes\n");    }else{        printf("No\n");    }    for(int i=b_len-1;i>=0;i--){        printf("%d",b[i]);    }    printf("\n");    return 0;   }