不得不承认智商真的被压制了。。
其实开始的时候试过用二项式定理避开组合数的计算，不过没想到单位矩阵这个神奇的东西233。
设有矩阵A，B，E
E为单位矩阵。
设新得到的矩阵为C，那么矩阵C的第一行第一列即为答案。
#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <cmath>#include <stack>#include <map>#pragma comment(linker, "/STACK:1024000000")#define EPS (1e-8)#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3fusing namespace std;const int MAXN = 5;struct MAT{int row,col;LL mat[MAXN][MAXN];void Init(int R,int C,int val){row = R,col = C;for(int i = 1;i <= row; ++i)for(int j = 1;j <= col; ++j)mat[i][j] = (i == j ? val : 0);}MAT Multi(MAT c,LL MOD){MAT tmp;tmp.Init(this->row,c.col,0);int i,j,k;for(k = 1;k <= this->col; ++k)for(i = 1;i <= tmp.row; ++i)for(j = 1;j <= tmp.col; ++j)(tmp.mat[i][j] += (this->mat[i][k]*c.mat[k][j])%MOD)%=MOD;return tmp;}MAT Quick(int n,LL MOD){MAT res,tmp = *this;res.Init(row,col,1);while(n){if(n&1)res = res.Multi(tmp,MOD);tmp = tmp.Multi(tmp,MOD);n >>= 1;}return res;}void Output(){cout<<"****************"<<endl;int i,j;for(i = 1;i <= row; ++i){for(j = 1;j <= col; ++j)printf("%3d ",mat[i][j]);puts("");}cout<<"&&&&&&&&&&&&&"<<endl;}};int main(){int n,m;MAT A,B;int T;scanf("%d",&T);while(T–){scanf("%d %d",&n,&m);A.Init(2,2,0);A.mat[1][1] = 1;A.mat[1][2] = 1;A.mat[2][1] = 1;A.mat[2][2] = 2;A = A.Quick(n,m);B.Init(2,1,0);B.mat[1][1] = 0;B.mat[2][1] = 1;B = A.Multi(B,m);printf("%I64d\n",B.mat[1][1]);}return 0;}
，当你能飞的时候就不要放弃飞
