2198 - 最简单的难题
我们定义F(0) = 7， F(1) = 11， F(n) = F(n - 1) + F(n - 2)
Input
输入数据包括多行，每行一个个数字n（n < 1,000,000）
Output
如果F(n)能被3整除就输出yes，不能则输出no
Examples
Input
0 1 2 3 4 5
Output
no no yes no no no
Solution C
#include<stdio.h> int main() { int n, i, a, b, c; while(scanf("%d",&n) != EOF) { a = 1; b = 2; c = 3; for(i = 2; i <= n; i++) { c = (a + b) % 3; a = b; b = c; } if(c == 0) puts("yes"); else puts("no"); } return 0; }
Solution C++
#include <cstdio> const int V = 1000000 + 50; int num[V]; int main() { int i, n; num[0] = 1; num[1] = 2; for(i = 2; i < V; ++i) num[i] = (num[i - 1] + num[i - 2]) % 3; while(~scanf("%d", &n)) printf("%s\n", num[n] ? "no" : "yes"); }