[Lonlife1031]Bob and Alice are eating food（递推，矩阵快速幂）

题目链接：http://www.ifrog.cc/acm/problem/1031 

题意：6个水果中挑出n个，使得其中2个水果个数必须是偶数，问有多少种选择方法。

设中0代表偶数，1代表奇数。分别代表两种水果的奇偶情况，有如下递推式：

 初始化的矩阵为：

） 以后写题解就用latex编辑公式了QAQ

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const LL mod = (LL)1e9+7;
const int maxn = 5;
LL n;
typedef struct Matrix {
    LL m[maxn][maxn];
    int r;
    int c;
    Matrix(){
        r = c = 0;
        memset(m, 0, sizeof(m));
    } 
} Matrix;
Matrix mul(Matrix m1, Matrix m2) {
    Matrix ans = Matrix();
    ans.r = m1.r;
    ans.c = m2.c;
    for(int i = 1; i <= m1.r; i++) {
        for(int j = 1; j <= m2.r; j++) {
               for(int k = 1; k <= m2.c; k++) {
                if(m2.m[j][k] == 0) continue;
                ans.m[i][k] = ((ans.m[i][k] + m1.m[i][j] * m2.m[j][k] % mod) % mod) % mod;
            }
        }
    }
    return ans;
}

Matrix quickmul(Matrix m, LL n) {
    Matrix ans = Matrix();
    for(int i = 1; i <= m.r; i++) {
        ans.m[i][i]  = 1;
    }
    ans.r = m.r;
    ans.c = m.c;
    while(n) {
        if(n & 1) {
            ans = mul(m, ans);
        }
        m = mul(m, m);
        n >>= 1;
    }
    return ans;
}

int main() {
    // freopen("in", "r", stdin);
    int T, _ = 1;
    scanf("%d", &T);
    while(T--) {
        scanf("%lld",&n);
        Matrix a; a.r = 4; a.c = 1;
        a.m[1][1] = 4; a.m[2][1] = 1; a.m[3][1] = 1; a.m[4][1] = 0;
        Matrix p; p.r = 4, p.c = 4;
        p.m[1][1] = 4; p.m[1][2] = 1; p.m[1][3] = 1; p.m[1][4] = 0;
        p.m[2][1] = 1; p.m[2][2] = 4; p.m[2][3] = 0; p.m[2][4] = 1;
        p.m[3][1] = 1; p.m[3][2] = 0; p.m[3][3] = 4; p.m[3][4] = 1;
        p.m[4][1] = 0; p.m[4][2] = 1; p.m[4][3] = 1; p.m[4][4] = 4;
        Matrix q = quickmul(p, n-1);
        Matrix b = mul(q, a);
        printf("Case #%d: %lld
", _++,b.m[1][1]);
    }
    return 0;
}