GCD

http://acm.hzau.edu.cn/problem.php?id=1202&csrf=gsbkpVmkV0QSB7bF1ZZYIdYM5y1coHa9

时间限制: 1 Sec  内存限制: 1280 MB

题目描述

输入

 The first line is an positive integer  T . (1<=T<= 10^3) indicates the number of test cases. In the next T lines, there are three positive integer n, m, p (1<= n,m,p<=10^9) at each line.

输出

样例输入

1 
1 2 3

样例输出

1

提示

gcd(1 + Sn, 1 + Sm) = gcd(fib(n + 2), fib(m + 2)) = fib(gcd(n + 2, m + 2)).

#include 
int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}
int main()
{
    int t, n, m, p, s, a, b, c;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d%d", &n, &m, &p);
        s = gcd(n + 2, m + 2);
        a = 1;b = 1;c = 1;
        for (int i = 3; i <= s; i++)
        {
            c = (a + b) % p;
            a = b;
            b = c;
        }
        printf("%d\n", c);
    }
    return 0;
}