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

Source

Andrew Stankevich's Contest #8

解题：坑爹的线段树

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <cmath>

 #include <algorithm>

 #include <climits>

 #include <vector>

 #include <queue>

 #include <cstdlib>

 #include <string>

 #include <set>

 #include <stack>

 #define LL long long

 #define pii pair<int,int>

 #define INF 0x3f3f3f3f

 using namespace std;

 const int maxn = ;

 struct Matrix {

     int m[][];

 };

 struct node {

     int lt,rt;

     Matrix matrix;

 };

 node tree[maxn<<];

 int r,n,m;

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

     Matrix c;

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

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

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

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

     return c;

 }

 void read(Matrix &c) {

     for(int i = ; i < ; ++i)

         for(int j = ; j < ; ++j)

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

 }

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

     tree[v].lt = lt;

     tree[v].rt = rt;

     if(lt == rt) {

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

         else {

             tree[v].matrix.m[][] = ;

             tree[v].matrix.m[][] = ;

             tree[v].matrix.m[][] = ;

             tree[v].matrix.m[][] = ;

         }

         return;

     }

     int mid = (lt+rt)>>;

     build(lt,mid,v<<);

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

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

 }

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

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

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

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

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

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

 }

 int main() {

     bool cao = false;

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

         int k = log2(n) + ;

         k = <<k;

         build(,k,);

         if(cao) puts("");

         cao = true;

         bool flag = false;

         while(m--){

             int s,t;

             if(flag) puts("");

             flag = true;

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

             Matrix ans = query(s,t,);

             for(int i = ; i < ; ++i)

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

         }

     }

     return ;

 }