SDNUOJ 1683.easy problem Ⅳ

Description

定义斐波那契数列，F(1) = F(2) = 1，对于所有的N，满足F(N+2)=F(N+1)+F(N)，求F(m)与F(n)的最大公约数
多组测试样例

Input

多组输入，每组两个整数，n, m(1 <= n <= 1000, 1 <= m <= 1000)

Output

F(m)与F(n)的最大公约数,结果不超过2⁶³-1

Sample Input

1 2

Sample Output

1

水题，斐波那契数列性质，注意开long long。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <sstream>
#include <cctype>
using namespace std;

typedef long long ll;
const ll mod = 1e9+7;

ll fib[100];

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a%b);
}

int main()
{
    fib[1] = fib[2] = 1;
    for(int i = 3; i <= 100; ++i)
        fib[i] = fib[i-1] + fib[i-2];
    ll n, m;
    while(cin >> n >> m)
        cout << fib[gcd(n, m)] << '\n';
	return 0;
}