Zju 2671 Cryptography

Cryptographytime limit: 5000 msmemory limit: 32768 kbthis 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 specified ted by his friend Andie. To do this, he must make some linear transformations over the ringZR = z/Rz.

Each linear transformation is defined2 × 2Matrix. Georgie has a sequence of MatricesA1, A2,...,. As a step of his algorithm he must take some segmentAI, AI + 1,..., AJOf the sequence and multiply some vector by a productPi, j = ai x ai + 1 X... x AJOf the segment. He must do itMVarious segments.

Help Georgie to determine the products he needs.

Inputthere are several test cases in the input. The first line of each case contains R( 1 & lt; = R & lt; = 10,000), N( 1 <= n <= 30,000) And M( 1 <= m <= 30,000). Next NBlocks of two lines, containing two integer numbers ranging from 0To R-1Each, describe matrices. blocks are separated with blank lines. They are followed MPairs of integer numbers ranging from 1To NEach that describe segments, products for which are to be calculated.
There is an empty line between cases. Output

PrintMBlocks containing two lines each. Each line shoshould contain two integer numbers ranging from0ToR-1And define the corresponding product matrix.
There shoshould be an empty line between cases.

Separate blocks with an empty line.

 

Sample

Input

3 4 40 10 02 11 20 00 21 00 21 42 31 32 2

Output

0 20 00 20 10 10 02 11 2

Sourceandrew stankevich's contest #8

Problem-solving: Line Segment tree

 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 long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 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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Zju 2671 Cryptography