Problem Description
There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).
Input
Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).
Output
Print the word "yes" if 3 divide evenly into F(n).
Print the word "no" if not.
Print the word "no" if not.
Sample Input
0 1 2 3 4 5
Sample Output
no#include<stdio.h> int f[1000010]; int main() { int n,i; while(~scanf("%d",&n)) { f[0]=7;f[1]=11; // if(n==0||n==1) // { // printf("no\n"); // continue; // } for(i=2;i<=n;i++) { f[i-1]=f[i-1]%3; f[i-2]=f[i-2]%3; f[i]=(f[i-1]+f[i-2])%3; } //printf("%d\n",f[n]); if(f[n]==0) printf("yes\n"); else printf("no\n"); } return 0; }
noyesnonono
Author
Leojay
分析:若直接求最后在求余,很明显这种类型的数列到40就会超出范围,故分别求余;(中国剩余定理)
代码: