题意：给出n个2*2的矩阵，有m个查询，问某段区间的矩阵乘积。
由于矩阵乘积符合结合律，故可用线段树来完成。
View Code
1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include <cstdlib> 7 #include <iostream> 8 #include <cmath> 9 #include <cstring> 10 #include <algorithm> 11 #include <string> 12 #include <vector> 13 #include <queue> 14 #include <stack> 15 #include <map> 16 #include <set> 17 using namespace std; 18 19 typedef long long ll; 20 const int INF = 2147483647; 21 22 /* 23 a b 24 c d 25 */ 26 #define lson l , m , rt << 1 27 #define rson m + 1 , r , rt << 1 | 1 28 29 struct Ma 30 { 31 int a,b,c,d; 32 }f[30005]; 33 34 int N,M,R; 35 Ma product(Ma A,Ma B) 36 { 37 Ma ret; 38 ret.a = (A.a*B.a+A.b*B.c)%R; 39 ret.b = (A.a*B.b+A.b*B.d)%R; 40 ret.c = (A.c*B.a+A.d*B.c)%R; 41 ret.d = (A.c*B.b+A.d*B.d)%R; 42 return ret; 43 } 44 Ma sum[30005<<2]; 45 void PushUp(int rt) 46 { 47 sum[rt] = product(sum[rt<<1],sum[rt<<1|1]); 48 } 49 void build(int l,int r,int rt) 50 { 51 if(l == r) 52 { 53 scanf("%d%d%d%d",&sum[rt].a,&sum[rt].b,&sum[rt].c,&sum[rt].d); 54 return ; 55 } 56 int m = (l + r) >> 1; 57 build(lson); 58 build(rson); 59 PushUp(rt); 60 } 61 Ma query(int L,int R,int l,int r,int rt) 62 { 63 if(L <= l && r <= R) 64 { 65 return sum[rt]; 66 } 67 int m = (l + r) >> 1; 68 Ma ret={1,0,0,1}; 69 if(L <= m)ret = product(ret,query(L,R,lson)); 70 if(m < R)ret = product(ret,query(L,R,rson)); 71 return ret; 72 } 73 int main() 74 { 75 // freopen("in","r",stdin); 76 bool flag=0; 77 while(~scanf("%d%d%d",&R,&N,&M)) 78 { 79 build(1,N,1); 80 while(M--) 81 { 82 if(flag)puts(""); 83 else flag=1; 84 int u,v; 85 scanf("%d%d",&u,&v); 86 Ma ret=query(u,v,1,N,1); 87 printf("%d %d\n%d %d\n",ret.a,ret.b,ret.c,ret.d); 88 } 89 } 90 return 0; 91 }