题目

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 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

背景

额，这道题在我大一时就遇到了。当时才学c语言，用的是笨办法，结果是部分正确。由于考研需要，近期重返了c语言，虽然二年没摸c语言了，但是由于大一时苦练，很快就拾起来了。现在看到这个题目时，觉得当年的代码写的跟**一样。不过我也感谢大一时的坚持。很快，我对这题就有了新的思路。

新的思路

总体思路

统计乘积前和乘积后序列各数字的频度，然后进行比较频度即可。

细说：

由于数字只有0-9，且序列长度不会超过20。所以用一有10个有效字符的字符数组存储频度即可。这样空间复杂度为O(1)。时间复杂度为O(n)。

#include"stdio.h"
#include"string.h"
int main(){
    char array[21]="\0";//保存原序列，经算法后，保存的是乘积后的序列。（若最高位溢出，不保存溢出位1。）
    gets(array);
    char freq[2][11]={"\0","\0"};//一维保存原序列个数字出现的频度，二维保存乘积后各数字出现的频度
    int length=strlen(array);
    int j=0;//工具
    int overFlow=0;//某位乘积加上低位进位后是否溢出。0位没溢出，1位溢出。
    int i=length - 1;
    while(i>=0){
        j=array[i]-'0';
        freq[0][j]+=1;//原序列 相应数字频度+1
        j=j*2;//单位数乘2
        if(overFlow==1){
            j++;
            overFlow=0;
        }
        //处理进位。找到不为9的前一位
        if(j>9){
            j=j%10;//产生乘积后的个位数字。
            overFlow=1;
        }
        freq[1][j]+=1;//乘积后个位数字的频度加1。有了上面if，保证了0=<j<=9
        array[i]=j+'0';
        i--;
    }

    //对比各数字频度是否相等
    for(int x=0;x<10;x++){
        if(freq[0][x]!=freq[1][x]){
            printf("No\n");
            break;
        }
        if(x==9)
            printf("Yes\n");
    }
    if(overFlow==1){
        printf("1");
    }
    puts(array);
    return 0;
}