zoj 2671 Cryptography（矩阵+线段树）

题意：给出n个2*2的矩阵，有m个查询，问某段区间的矩阵乘积。

由于矩阵乘积符合结合律，故可用线段树来完成。

/*
Author:Zhaofa Fang
Lang:C++
*/
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;

typedef long long ll;
const int INF = 2147483647;

/*
    a b
    c d
*/
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1

struct Ma
{
    int a,b,c,d;
}f[30005];

int N,M,R;
Ma product(Ma A,Ma B)
{
    Ma ret;
    ret.a = (A.a*B.a+A.b*B.c)%R;
    ret.b = (A.a*B.b+A.b*B.d)%R;
    ret.c = (A.c*B.a+A.d*B.c)%R;
    ret.d = (A.c*B.b+A.d*B.d)%R;
    return ret;
}
Ma sum[30005<<2];
void PushUp(int rt)
{
    sum[rt] = product(sum[rt<<1],sum[rt<<1|1]);
}
void build(int l,int r,int rt)
{
    if(l == r)
    {
        scanf("%d%d%d%d",&sum[rt].a,&sum[rt].b,&sum[rt].c,&sum[rt].d);
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}
Ma query(int L,int R,int l,int r,int rt)
{
    if(L <= l && r <= R)
    {
        return sum[rt];
    }
    int m = (l + r) >> 1;
    Ma ret={1,0,0,1};
    if(L <= m)ret = product(ret,query(L,R,lson));
    if(m < R)ret = product(ret,query(L,R,rson));
    return ret;
}
int main()
{
 //   freopen("in","r",stdin);
    bool flag=0;
    while(~scanf("%d%d%d",&R,&N,&M))
    {
        build(1,N,1);
        while(M--)
        {
            if(flag)puts("");
            else flag=1;
            int u,v;
            scanf("%d%d",&u,&v);
            Ma ret=query(u,v,1,N,1);
            printf("%d %d\n%d %d\n",ret.a,ret.b,ret.c,ret.d);
        }
    }
    return 0;
}