hdoj 4565 So Easy!（构造无理数共轭，矩阵快速幂）

最难的是要想到无理数共轭...

还要注意注意负数的取模，可以先+mod再进行取模.

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a, b, m;
struct node
{
    ll s[2][2];
    node() {}
    node(ll a, ll b, ll c, ll d)
    {
        s[0][0] = a;
        s[0][1] = b;
        s[1][0] = c;
        s[1][1] = d;
    }
};

node mul(node a, node b)
{
    node t = node(0, 0, 0, 0);
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
            for(int k = 0; k < 2; k++)
                t.s[i][j] = (t.s[i][j]+a.s[i][k]*b.s[k][j]+m)%m;    //注意负数
    return t;
}

node mt_pow(node p, int n)
{
    node q = node(1, 0, 0, 1);
    while(n)
    {
        if(n&1) q = mul(p, q);
        p = mul(p, p);
        n /= 2;
    }
    return q;
}

int main(void)
{
    ll n;
    while(cin >> a >> b >> n >> m)
    {
        node base = node(2*a, (b-a*a), 1, 0);
        ll c1 = 2*a%m;
        ll c2 = 2*(a*a+b)%m;
        if(n == 1) printf("%lld\n", c1);
        else if(n == 2) printf("%lld\n", c2);
        else
        {
            node ans = mt_pow(base, n-2);
            printf("%lld\n", (ans.s[0][0]*c2+ans.s[0][1]*c1+m)%m);
        }
    }
    return 0;
}

So Easy!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3883 Accepted Submission(s): 1279

Problem Description

　　A sequence S _n is defined as:

Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate S _n.
　　You, a top coder, say: So easy!

Input

　　There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 2 ¹⁵, (a-1) ²< b < a ², 0 < b, n < 2 ³¹.The input will finish with the end of file.

Output

　　For each the case, output an integer S _n.

Sample Input