LightOJ 1065

http://www.lightoj.com/volume_showproblem.php?problem=1065

题意：给出递推式f(0) = a, f(1) = b, f(n) = f(n - 1) +f(n - 2) 求f(n)

思路：给出了递推式就是水题。

/** @Date    : 2016-12-17-15.54
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version :
  */
#include<bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;
struct matrix
{
    LL mat[2][2];
    void init()
    {
        mat[0][0] = mat[1][0] = mat[0][1] = mat[1][1] = 0;
    }
};

matrix mul(matrix a, matrix b, LL mod)
{
    matrix c;
    c.init();
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
            for(int k = 0; k < 2; k++)
            {
                c.mat[i][j] += a.mat[i][k] * b.mat[k][j];
                c.mat[i][j] %= mod;
            }
    return c;
}

matrix fpow(matrix x, LL n, LL mod)
{
    matrix r;
    r.init();
    for(int i = 0; i < 2; i++)
        r.mat[i][i] = 1;
    while(n > 0)
    {
        if(n & 1)
            r = mul(r, x, mod);
        x = mul(x, x, mod);
        n >>= 1;
    }
    return r;
}

LL Tis(LL x, LL a, LL b, LL mod)
{
    matrix t;
    t.init();
    t.mat[0][0] = 1;
    t.mat[0][1] = 1;
    t.mat[1][0] = 1;
    matrix s;
    s = fpow(t, x - 1, mod);
    LL ans = s.mat[0][0]*b + s.mat[1][0]*a;
    return ans % mod;
}
int main()
{
    int T;
    cin >> T;
    int cnt = 0;
    while(T--)
    {
        LL n, a, b, m;
        scanf("%lld%lld%lld%lld", &a, &b, &n, &m);
        LL x = 1;
        while(m--)
        {
            x*=10;
        }
        printf("Case %d: %lld
",++cnt, Tis(n, a, b, x));
    }
    return 0;
}