UVA10689 Yet another Number Sequence —— 斐波那契、矩阵快速幂

题目链接：https://vjudge.net/problem/UVA-10689

题解：

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
//const int MOD = 1e9+7;
const int MAXN = 1e6+100;

LL MOD;
const int Size = 2;
struct MA
{
    LL mat[Size][Size];
    void init()
    {
        for(int i = 0; i<Size; i++)
        for(int j = 0; j<Size; j++)
            mat[i][j] = (i==j);
    }
};

MA mul(MA x, MA y)
{
    MA ret;
    memset(ret.mat, 0, sizeof(ret.mat));
    for(int i = 0; i<Size; i++)
    for(int j = 0; j<Size; j++)
    for(int k = 0; k<Size; k++)
        ret.mat[i][j] += (1LL*x.mat[i][k]*y.mat[k][j])%MOD, ret.mat[i][j] %= MOD;
    return ret;
}

MA qpow(MA x, LL y)
{
    MA s;
    s.init();
    while(y)
    {
        if(y&1) s = mul(s, x);
        x = mul(x, x);
        y >>= 1;
    }
    return s;
}

MA tmp = {
    1, 1,
    1, 0
};

int main()
{
    LL f[2], n, m;
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld%lld%lld%lld",&f[0],&f[1],&n,&m);
        if(n<2)
        {
            printf("%lld
", f[n]);
            continue;
        }

        MOD = 1;
        while(m--) MOD *= 10;
        MA s = tmp;
        s = qpow(s, n-1);
        LL ans = (1LL*s.mat[0][0]*f[1]%MOD+1LL*s.mat[0][1]*f[0]%MOD)%MOD;
        printf("%lld
", ans);
    }
}