hdu 1021 - Fibonacci Again
2013-09-20 14:12
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题目:Fibonacci Again
题意:有一个新的斐波那契数列:f[0]=7,f[1]=11 , f
= 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:
题意:有一个新的斐波那契数列:f[0]=7,f[1]=11 , f
= 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 ? "no" : "yes") << endl; return 0; }
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