Cryptography
Time Limit: 5000ms
Memory Limit: 32768KB
This problem will be judged on ZJU. Original ID: 267164-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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