又是斐波那契数列-数学
Time Limit: 1000MS | Memory Limit: 65535KB | 64bit IO Format: %I64d & %I64u |
Description
有另一种斐波那契数列:F(0)=7,F(1)=11,F(n)=F(n-1)+F(n-2) (n>=2)
Input
输入数据有多行组成,每一行上是一个整数n(n<1000000);
Output
如果F(n)能被3整除,那么打印一行"yes";否则,打印一行"no".
Sample Input
0 1 2 3 4 5
Sample Output
no no yes no no no
Source
Timebug
1 #include<iostream> 2 #include<algorithm> 3 #include<queue> 4 #include<cstdio> 5 #include<cstdlib> 6 #include<cstring> 7 #include<cmath> 8 using namespace std; 9 #define sr(x) scanf("%d",&x) 10 #define sc(x) printf("%d",x) 11 #define hh printf("\n") 12 #define mod 2011 13 int main() 14 { 15 int n; 16 while(sr(n)!=EOF) 17 { 18 n%=8; 19 if(n==2||n==6)printf("yes\n"); 20 else printf("no\n"); 21 } 22 return 0; 23 }