自测-4 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 kk 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

说明：最开始使用的思路是大数计算的方式，存入字符串数组，然后运算后与原数组进行对比，但存在wa点，现在没找到。先放置于此：

#include <stdio.h>
#include<string.h>
int main()
{
    // N为长度，M为进位
    int N = ,M = , i = , num = ,loop = , j = ;
    char str[];
    char tmp[];
    scanf("%s",str);
    N = strlen(str);

    for(i = N - ; i >= ; i--){
        num = (str[i] - '')* + M;
        M = ;
        if(num > ){
            num = num % ;
            M++;
        }
        tmp[i] = num + '';
    }
    for(i = ; i < N; i++){
        for(j = ; j < N; j++){
            if(tmp[i] == str[j]){
                str[j] = 'a';
                break;
            }
        }
    }
    for(i = ; i < N; i++){
        if(str[i] != 'a'){
            loop = ;
            break;
        }
    }
    if(M == ){
        printf("No\n1");
        for(i = ; i < N; i++){
            printf("%d",tmp[i] - '');
        }
        return ;
    }
    if(loop == ){
        for(i = ; i < N; i++){
            printf("%d",tmp[i] - '');
        }
    }else{
        printf("Yes\n");
        for(i = ; i < N; i++){
            printf("%d",tmp[i] - '');
        }
    }
    return ;
}

后改变思路，只记录0~9十个数组出现的次数，仅需要两个长度为10的一维数组即可。使用别人写的C++代码，引用链接在最后

 #include <cstdio>
 #include <string>
 #include <iostream>
 using namespace std;
 int cnt1[] = {  }, cnt2[] = {};  

 int main(){
     string s;
     cin >> s;
     string s2 = s;
     for (int i = ; i < s.size(); i++){
         cnt1[s[i] - '']++;
     }
     int carry = ;
     for (int i = s.size() - ; i >= ; i--){
         s2[i] = ((s[i] - '') *  + carry )%  + '';
         carry = ((s[i] - '') *  + carry) / ;
     }
     if (carry){
         char c = carry + '';
         s2 = c + s2;
         cout << "No\n" << s2 << endl;
         return ;
     }
     for (int i = ; i < s2.size(); i++){
         cnt2[s2[i] - '']++;
     }
     bool flag = true;
     for (int i = ; i < ; i++){
         if (cnt1[i] != cnt2[i]){
             flag = false;
             break;
         }
     }
     if (flag) cout << "Yes" << "\n" << s2 << endl;
     else cout << "No\n" << s2;
     return ;
 }  

http://blog.csdn.net/xtzmm1215/article/details/38837203