[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+;
 const int maxn = ;
 LL n;
 typedef struct Matrix {
     LL m[maxn][maxn];
     int r;
     int c;
     Matrix(){
         r = c = ;
         memset(m, , sizeof(m));
     }
 } Matrix;
 Matrix mul(Matrix m1, Matrix m2) {
     Matrix ans = Matrix();
     ans.r = m1.r;
     ans.c = m2.c;
     for(int i = ; i <= m1.r; i++) {
         for(int j = ; j <= m2.r; j++) {
                for(int k = ; k <= m2.c; k++) {
                 if(m2.m[j][k] == ) 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 = ; i <= m.r; i++) {
         ans.m[i][i]  = ;
     }
     ans.r = m.r;
     ans.c = m.c;
     while(n) {
         if(n & ) {
             ans = mul(m, ans);
         }
         m = mul(m, m);
         n >>= ;
     }
     return ans;
 }

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