ZOJ 2671 Cryptography（线段树+求区间矩阵乘积）

题意：已知n个矩阵（下标从1开始），求下标x~y区间矩阵的乘积。最多m次询问，n ( 1 <= n <= 30,000) and m ( 1 <= m <= 30,000)。

分析：

1、矩阵初始化为单位矩阵，因为要做乘积，E*A=A。

2、因为输出矩阵的所有值范围在0~r-1，所以要对r取余。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e4;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 30000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
struct Matrix{
    int ma[2][2];
    Matrix(){
        ma[0][0] = ma[1][1] = 1;
        ma[0][1] = ma[1][0] = 0;
    }
}num[MAXN << 2];
int r;
Matrix mul(Matrix a, Matrix b){
    Matrix ans;
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            ans.ma[i][j] = 0;
            for(int k = 0; k < 2; ++k){
                (ans.ma[i][j] += (a.ma[i][k] * b.ma[k][j]) % r) %= r;
            }
        }
    }
    return ans;
}
void add_up(int id){
    num[id] = mul(num[id << 1], num[id << 1 | 1]);
}
void add(int l, int r, int cur, Matrix x, int id){
    if(l == r){
        num[id] = x;
        return;
    }
    int mid = (l + r) >> 1;
    if(cur <= mid){
        add(l, mid, cur, x, id << 1);
    }
    else add(mid + 1, r, cur, x, id << 1 | 1);
    add_up(id);
}
Matrix query(int L, int R, int l, int r, int id){
    if(l >= L && r <= R){
        return num[id];
    }
    int mid = (l + r) >> 1;
    Matrix ans;
    if(L <= mid){
        ans = mul(ans, query(L, R, l, mid, id << 1));
    }
    if(R >= mid + 1){
        ans = mul(ans, query(L, R, mid + 1, r, id << 1 | 1));
    }
    return ans;
}
int main(){
    int n, m;
    bool flag = true;
    while(scanf("%d%d%d", &r, &n, &m) == 3){
        if(flag) flag = false;
        else printf("\n");
        for(int i = 1; i <= n; ++i){
            Matrix tmp;
            for(int j = 0; j < 2; ++j){
                for(int k = 0; k < 2; ++k){
                    scanf("%d", &tmp.ma[j][k]);
                }
            }
            add(1, n, i, tmp, 1);
        }
        while(m--){
            int x, y;
            scanf("%d%d", &x, &y);
            Matrix t = query(x, y, 1, n, 1);
            printf("%d %d\n", t.ma[0][0], t.ma[0][1]);
            printf("%d %d\n", t.ma[1][0], t.ma[1][1]);
            if(m) printf("\n");
        }
    }
    return 0;
}