题意:有一个新的斐波那契数列:f[0]=7,f[1]=11 , f[n] = f[n-1]+f[n-2] (n >= 2) 求第n项能否被3整除
思路:同余定理,水题中的战斗机不解释
代码:
#pragma comment(linker, "/STACK:102400000,102400000")
#include "iostream"
#include "cstring"
#include "algorithm"
#include "cmath"
#include "cstdio"
#include "sstream"
#include "queue"
#include "vector"
#include "string"
#include "stack"
#include "cstdlib"
#include "deque"
#include "fstream"
#include "map"
using namespace std;
typedef long long LL;
const int INF = 0x1fffffff;
const int MAXN = 1000000+100;
#define eps 1e-14
int fb[1000000+10];
void mk()
{
fb[0] = 7%3;
fb[1] = 11%3;
for(int i = 2 ; i < 1000000+10 ; i++)
fb[i] = (fb[i-1]%3 + fb[i-2]%3)%3;
}
int main()
{
int n;
mk();
while(cin >> n)
cout << (fb[n] ? "no" : "yes") << endl;
return 0;
}