zoj 2671 Cryptography（矩阵+线段树）

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 }