7-49 Have Fun with Numbers（20 分）

题目：（大整数用字符数组存储）

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

代码：

#include <stdio.h>
#include <string.h>
#define N 21

int main(){
    char original[N];
    int doubledNum[N], i, digitsCount[10];
    int carry=0, isSame=1; //记录原始大数乘以2后是否产生进位
    gets(original);

    for(i=0;i<strlen(original);i++){ //得到每位乘2的未处理进位的数
        doubledNum[i] = 2*(original[i]-'0');
    }
    for(i=strlen(original)-1;i>=0;i--){
        if(i>0){
            doubledNum[i-1] += doubledNum[i]/10;
            doubledNum[i] %= 10;
        }
        else{
            carry += doubledNum[i]/10;
            doubledNum[i] %= 10;
        }

    }
    if(carry){
        isSame = 0;
    }
    else{
        for(i=0;i<10;i++){ //初始化10个位值的计数数组
            digitsCount[i] = 0;
        }
        for(i=0;i<strlen(original);i++){ //逐位增加计数
            digitsCount[original[i]-'0']++;
        }
        for(i=0;i<strlen(original);i++){ //逐位消减计数
            digitsCount[doubledNum[i]]--;
        }
        for(i=0; i<10; i++){ //检查各个位值的计数
            if(digitsCount[i]!=0){
                isSame = 0;
                break;
            }
        }
    }
    if(isSame){
        printf("Yes\n");
    }
    else{
        printf("No\n");
    }
    if(carry){
        printf("%d", carry);
    }
    for(i=0;i<strlen(original);i++){
        printf("%d", doubledNum[i]);
    }

    return 0;
}