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题目解析
本题给出一个长度再20以内的数字（由1~9组成），要求判断这个数字加倍后的新数字是不是这个数字的某一种排列，如果是的化输出Yes否则输出No，之后输出加倍后的数字。
由于数字最大位数为20位，超过了long long int的记录范围我们用数组num记录这个数字，用数组cnt记录num中1~9出现的次数，将num加倍后判断其中1~9出现的次数是否发送改变，若没有发送改变则证明加倍后的数字是原数字的某一种排列，反之则不是。
AC代码
1 #include <bits/stdc++.h>
2 using namespace std;
3 int num[22];
4 int cnt[10];
5 string str;
6 void toInt(){
7 for(int i = 0; i < str.size(); i++)
8 num[i] = str[i] - ‘0‘;
9 }
10 void getCnt(){
11 for(int i = 0; i < str.size(); i++)
12 cnt[num[i]]++;
13 }
14 bool judge(int carry){
15 if(carry != 0) //如果最高位进位不为零，则证明加倍后的数字比原数字多一位，那么其肯定不是原数字的一个排列
16 return false;
17 for(int i = 0; i < str.size(); i++)
18 cnt[num[i]]--;
19 for(int i = 1; i <= 9; i++){ //判断新的num中1~9的数量是否和加倍前一样
20 if(cnt[i] != 0)
21 return false;
22 }
23 return true;
24 }
25 int doubleNumber(){ //将数组num加倍并返回最高位进位
26 int carry = 0;
27 for(int i = str.size() - 1; i >= 0; i--){
28 int temp = num[i];
29 num[i] = (2 * temp + carry) % 10;
30 carry = 2 * temp / 10;
31 }
32 return carry;
33 }
34 int main()
35 {
36 cin >> str; //输入数字
37 toInt(); //将输入的数字转化为数组
38 getCnt(); //获取数组中1~9出现的次数
39 int carry = doubleNumber(); //将num加倍carry记录最高位的进位
40 if(judge(carry)){ //判断加倍后的数字是否为原数字的某一个排列
41 printf("Yes\n");
42
43 }else
44 printf("No\n");
45 if(carry != 0) //判断是否需要输出进位
46 printf("%d", carry);
47 for(int i = 0; i < str.size(); i++) //输出加倍后的数组num
48 printf("%d", num[i]);
49 printf("\n");
50 return 0;
51 }
原文地址：https://www.cnblogs.com/suvvm/p/10311268.html