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 = 0,M = 0, i = 0, num = 0,loop = 0, j = 0;
    char str[100];
    char tmp[100];
    scanf("%s",str);
    N = strlen(str);

    for(i = N - 1; i >= 0; i--){
        num = (str[i] - '0')*2 + M;
        M = 0;
        if(num > 9){
            num = num % 10;
            M++;
        }
        tmp[i] = num + '0';
    }
    for(i = 0; i < N; i++){
        for(j = 0; j < N; j++){
            if(tmp[i] == str[j]){
                str[j] = 'a';
                break;
            }
        }
    }
    for(i = 0; i < N; i++){
        if(str[i] != 'a'){
            loop = 1;
            break;
        }
    }
    if(M == 1){
        printf("No
1");
        for(i = 0; i < N; i++){
            printf("%d",tmp[i] - '0');
        }
        return 0;
    }
    if(loop == 1){
        for(i = 0; i < N; i++){
            printf("%d",tmp[i] - '0');
        }
    }else{
        printf("Yes
");
        for(i = 0; i < N; i++){
            printf("%d",tmp[i] - '0');
        }
    }
    return 0;
}

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[10] = { 0 }, cnt2[10] = {0};  
  
int main(){  
    string s;  
    cin >> s;  
    string s2 = s;  
    for (int i = 0; i < s.size(); i++){  
        cnt1[s[i] - '0']++;  
    }  
    int carry = 0;  
    for (int i = s.size() - 1; i >= 0; i--){  
        s2[i] = ((s[i] - '0') * 2 + carry )% 10 + '0';  
        carry = ((s[i] - '0') * 2 + carry) / 10;  
    }  
    if (carry){  
        char c = carry + '0';  
        s2 = c + s2;  
        cout << "No
" << s2 << endl;  
        return 0;  
    }  
    for (int i = 0; i < s2.size(); i++){  
        cnt2[s2[i] - '0']++;  
    }  
    bool flag = true;  
    for (int i = 0; i < 10; i++){  
        if (cnt1[i] != cnt2[i]){  
            flag = false;  
            break;  
        }  
    }  
    if (flag) cout << "Yes" << "
" << s2 << endl;  
    else cout << "No
" << s2;  
    return 0;  
}  

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