hdoj 1719 Friend
Friend
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2234 Accepted Submission(s): 1121
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!思路: 设c是Friend数,题中给出公式ab+a+b。则c=ab+a+b。c+1=ab+a+b+1=(a+1)*(b+1),所以推断n+1是否满足这个公式就可以。又由于1,2是最小的Friend数。所以推断n+1=(a+1)^x*(b+1)^y即n+1=2^x*3^y是否成立。
代码:#include<stdio.h> int main() { int n,x,y; while(scanf("%d",&n)!=EOF) { n+=1;//n先加1; x=0;y=0; while(n%2==0) { n=n/2; x++; } while(n%3==0) { n=n/3; y++; } if(n==1&&(x>0||y>0))//n=0时不是Friend数。 printf("YES!\n"); else printf("NO!\n"); } return 0; }