/*Fibonacci Again Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 44042 Accepted Submission(s): 21024 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. Sample Input 0 1 2 3 4 5 Sample Output no no yes no no no Author Leojay */ #include<stdio.h> int a[1000010]; int main() { int i, j, k; a[0] = 1; a[1] = 2; for(i = 2; i < 1000002; ++i) a[i] = (a[i-1] + a[i-2]) % 3; while(scanf("%d", &k) != EOF) if(a[k] == 0) printf("yes\n"); else printf("no\n"); return 0; }
题意:给出一个新的fib数列,给出一个n让你判断这个数是否是3的倍数。
思路:由于fib数列是累加的,所以我们可以用同于定理直接累加上去,即使n很大也不会超出范围。