Friend
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2131 Accepted Submission(s): 1074
Problem Description
Friend number are defined recursively as follows.
(1) numbers 1 and 2 are friend number;
(2) if a and b are friend numbers, so is ab+a+b;
(3) only the numbers defined in (1) and (2) are friend number.
Now your task is to judge whether an integer is a friend number.
(1) numbers 1 and 2 are friend number;
(2) if a and b are friend numbers, so is ab+a+b;
(3) only the numbers defined in (1) and (2) are friend number.
Now your task is to judge whether an integer is a friend number.
Input
There are several lines in input, each line has a nunnegative integer a, 0<=a<=2^30.
Output
For the number a on each line of the input, if a is a friend number, output “YES!”, otherwise output “NO!”.
Sample Input
3
13121
12131
Sample Output
YES!
YES!
NO!
Source
Recommend
friend数： n = (a + 1)(b + 1) - 1;
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 using namespace std; 5 int main(){ 6 int n; 7 while(~scanf("%d", &n)){ 8 if(!n){ 9 printf("NO! "); 10 continue; 11 } 12 n += 1; 13 while(n % 2 == 0 || n % 3 == 0){ 14 if(n % 2 == 0) 15 n /= 2; 16 else 17 n /= 3; 18 } 19 if(n == 1) 20 printf("YES! "); 21 else 22 printf("NO! "); 23 } 24 return 0; 25 }