ZJU 2671 Cryptography

Cryptography

Time Limit: 5000ms

Memory Limit: 32768KB

This problem will be judged on  ZJU. Original ID: 2671
64-bit integer IO format: %lld      Java class name: Main

 

Young cryptoanalyst Georgie is planning to break the new cipher invented by his friend Andie. To do this, he must make some linear transformations over the ring Zr = Z/rZ.

Each linear transformation is defined by 2×2 matrix. Georgie has a sequence of matrices A1 , A2 ,..., An . As a step of his algorithm he must take some segment Ai , Ai+1 , ..., Aj of the sequence and multiply some vector by a product Pi,j=Ai × Ai+1 × ... × Aj of the segment. He must do it for m various segments.

Help Georgie to determine the products he needs.

Input

There are several test cases in the input. The first line of each case contains r (1 <= r <= 10,000), n (1 <= n <= 30,000) and m (1 <= m <= 30,000). Next n blocks of two lines, containing two integer numbers ranging from 0 to r - 1 each, describe matrices. Blocks are separated with blank lines. They are followed by m pairs of integer numbers ranging from1 to n each that describe segments, products for which are to be calculated.
There is an empty line between cases.

Output

Print m blocks containing two lines each. Each line should contain two integer numbers ranging from 0 to r - 1 and define the corresponding product matrix.
There should be an empty line between cases.

Separate blocks with an empty line.

 

Sample

Input

3 4 4

0 1

0 0



2 1

1 2



0 0

0 2



1 0

0 2



1 4

2 3

1 3

2 2

Output

0 2

0 0



0 2

0 1



0 1

0 0



2 1

1 2

Source

Andrew Stankevich's Contest #8

 

解题：坑爹的线段树

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 40000;

18 struct Matrix {

19     int m[4][4];

20 };

21 struct node {

22     int lt,rt;

23     Matrix matrix;

24 };

25 node tree[maxn<<2];

26 int r,n,m;

27 Matrix Multiplication(const Matrix &a,const Matrix &b) {

28     Matrix c;

29     c.m[0][0] = (a.m[0][0]*b.m[0][0] + a.m[0][1]*b.m[1][0])%r;

30     c.m[0][1] = (a.m[0][0]*b.m[0][1] + a.m[0][1]*b.m[1][1])%r;

31     c.m[1][0] = (a.m[1][0]*b.m[0][0] + a.m[1][1]*b.m[1][0])%r;

32     c.m[1][1] = (a.m[1][0]*b.m[0][1] + a.m[1][1]*b.m[1][1])%r;

33     return c;

34 }

35 void read(Matrix &c) {

36     for(int i = 0; i < 2; ++i)

37         for(int j = 0; j < 2; ++j)

38             scanf("%d",&c.m[i][j]);

39 }

40 void build(int lt,int rt,int v) {

41     tree[v].lt = lt;

42     tree[v].rt = rt;

43     if(lt == rt) {

44         if(lt <= n) read(tree[v].matrix);

45         else {

46             tree[v].matrix.m[0][0] = 1;

47             tree[v].matrix.m[1][1] = 0;

48             tree[v].matrix.m[0][1] = 0;

49             tree[v].matrix.m[1][0] = 0;

50         }

51         return;

52     }

53     int mid = (lt+rt)>>1;

54     build(lt,mid,v<<1);

55     build(mid+1,rt,v<<1|1);

56     tree[v].matrix = Multiplication(tree[v<<1].matrix,tree[v<<1|1].matrix);

57 }

58 Matrix query(int lt,int rt,int v){

59     if(tree[v].lt == lt && tree[v].rt == rt) return tree[v].matrix;

60     int mid = (tree[v].lt + tree[v].rt)>>1;

61     if(rt <= mid) return query(lt,rt,v<<1);

62     else if(lt > mid) return query(lt,rt,v<<1|1);

63     else return Multiplication(query(lt,mid,v<<1),query(mid+1,rt,v<<1|1));

64 }

65 int main() {

66     bool cao = false;

67     while(~scanf("%d %d %d",&r,&n,&m)){

68         int k = log2(n) + 1;

69         k = 1<<k;

70         build(1,k,1);

71         if(cao) puts("");

72         cao = true;

73         bool flag = false;

74         while(m--){

75             int s,t;

76             if(flag) puts("");

77             flag = true;

78             scanf("%d %d",&s,&t);

79             Matrix ans = query(s,t,1);

80             for(int i = 0; i < 2; ++i)

81                 printf("%d %d\n",ans.m[i][0],ans.m[i][1]);

82         }

83     }

84     return 0;

85 }

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