PTA (Advanced Level)1023 Have Fun with Numbers

佚名 5年前 (2019-01-23) 算法 1364人围观抢沙发百度已收录

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.

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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 }