小米 OJ 编程比赛 02 月常规赛 3 Logic Gatekeeper CDQ分治

link：https://code.mi.com/problem/list/view?id=139

题意：

　　有一个1e6 * 1e6 大的格子，现在有两种操作：1，给一个子矩阵中的每个格子加上k。2，计算一个子矩阵中格子数字的和，在mod意义下除以子矩阵的大小。

思路：

　　首先要学一下（ http://www.cnblogs.com/RabbitHu/p/BIT.html）中关于二位矩阵区间修改，求区间和的知识，然后由于这个格子太大，我们就要用cdq分治降维。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
//#include <unordered_map>
/*

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*/

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);

const ll oo = 1ll<<;
const ll mos = 0x7FFFFFFF;  //
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = ;
const double esp = 1e-;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;

template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}

inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;}
#define MODmul(a, b) ((a*b >= mod) ? ((a*b)%mod + 2*mod) : (a*b))
#define MODadd(a, b) ((a+b >= mod) ? ((a+b)%mod + 2*mod) : (a+b))

/*-----------------------showtime----------------------*/
            const int maxn = 1e6+;
            struct node{
                int op,x,y;
                ll val;
            }a[maxn << ],tmp[maxn<<];

            int lowbit(int x){
                return x & (-x);
            }

            struct bit{
                ll sum[maxn];

                void add(ll x,ll c){
                    while(x < maxn){
                        sum[x] = ((sum[x] + c)%mod + mod)%mod;
                        x += lowbit(x);
                    }
                }
                ll getsum(int x){
                    ll res = ;
                    while(x > ) {
                        res = ((res + sum[x])% mod + mod)%mod;
                        x -= lowbit(x);
                    }
                    return res;
                }
            }A,B,C,D;

            queue<int>que;
            ll ans[maxn],sz[maxn];

            void update(int x,int y,ll val){
                A.add(y, (val%mod + mod )%mod);
                B.add(y, (val*x%mod + mod )% mod);
                C.add(y, (val*y%mod + mod) % mod);
                D.add(y, (val*x%mod*y%mod+mod)%mod);
            }

            ll solve(int x, int y){
                ll res = ;
                res = (res + 1ll*(x+) * (y+) % mod * A.getsum(y)%mod )%mod;
                res = (res - 1ll*(y+) * B.getsum(y))%mod;
                res = (res - 1ll*(x+) * C.getsum(y))%mod;
                res = (res +  D.getsum(y))%mod;
                res = (res + mod)%mod;
                return res;
            }
            void cdq(int le,int ri){
                if(le >= ri) return;
                int mid = (le + ri) >> ;
                cdq(le, mid); cdq(mid+, ri);

                int p = le,q = mid+;
                int tot = ;
                while(p <= mid && q <= ri) {
                    if(a[p].x <= a[q].x){
                        if(a[p].op == ) {
                            update(a[p].x, a[p].y,a[p].val);
                            que.push(p);
                        }
                        tmp[++tot] = a[p++];
                    }
                    else {
                        if(a[q].op == ) {
                            ans[a[q].val] = (ans[a[q].val] + solve(a[q].x, a[q].y) ) % mod;
                        }
                        else if(a[q].op == ) {
                            ans[a[q].val] = ((ans[a[q].val] - solve(a[q].x, a[q].y) ) ) % mod;
                            if(ans[a[q].val]< ) ans[a[q].val] = (ans[a[q].val]+mod)%mod;
                        }
                        tmp[++tot] = a[q++];
                    }
                }

                while(p <= mid) tmp[++tot] = a[p++];
                while(q <= ri){
                        if(a[q].op == ) {
                            ans[a[q].val] = (ans[a[q].val] + solve(a[q].x, a[q].y) ) % mod;
                        }
                        else if(a[q].op == ) {
                            ans[a[q].val] = ((ans[a[q].val] - solve(a[q].x, a[q].y) ) ) % mod;
                            if(ans[a[q].val]< ) ans[a[q].val] = (ans[a[q].val]+mod)%mod;
                        }
                        tmp[++tot] = a[q++];
                }
                while(!que.empty()) {
                    int p = que.front(); que.pop();
                    update(a[p].x, a[p].y,-1ll*a[p].val);
                }
                rep(i, , tot) a[i+le-] = tmp[i];
            }
            ll ksm(ll a, ll n){
                ll res = ;
                while(n > ){
                    if(n & ) res = res * a % mod;
                    a = a * a % mod;
                    n>>=;
                }
                return res;
            }
int main(){
            int n,m,q;
            scanf("%d%d%d", &n, &m, &q);
            int tot = ,id = ;
            while(q--) {
                int op; scanf("%d", &op);
                if(op == ) {
                    int x1,y1,x2,y2,k;
                    scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k);
                    tot++;  a[tot].x = x1; a[tot].y = y1; a[tot].val = k;a[tot].op = ;
                    tot++;  a[tot].x = x1; a[tot].y = y2+; a[tot].val = -k;a[tot].op = ;
                    tot++;  a[tot].x = x2+; a[tot].y = y1; a[tot].val = -k;a[tot].op = ;
                    tot++;  a[tot].x = x2+; a[tot].y = y2+; a[tot].val = k;a[tot].op = ;
                }
                else {
                    int x1,y1,x2,y2;
                    id++;
                    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                    tot++;  a[tot].x = x1-; a[tot].y = y1-;   a[tot].val = id; a[tot].op = ;
                    tot++;  a[tot].x = x1-; a[tot].y = y2; a[tot].val = id; a[tot].op = ;
                    tot++;  a[tot].x = x2; a[tot].y = y1-; a[tot].val = id; a[tot].op = ;
                    tot++;  a[tot].x = x2; a[tot].y = y2; a[tot].val = id;a[tot].op = ;
                    sz[id] = 1ll*(y2-y1+)*(x2-x1+)%mod;
                }
            }

            cdq(, tot);

            rep(i, , id) {
                printf("%lld\n", 1ll*ans[i] * ksm(sz[i], mod-)%mod);
            }
            return ;
}