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>

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<<17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 998244353;
const double esp = 1e-8;
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=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',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+9;
struct node{
    int op,x,y;
    ll val;
}a[maxn << 2],tmp[maxn<<2];

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 = 0;
        while(x > 0) {
            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 = 0;
    res = (res + 1ll*(x+1) * (y+1) % mod * A.getsum(y)%mod )%mod;
    res = (res - 1ll*(y+1) * B.getsum(y))%mod;
    res = (res - 1ll*(x+1) * 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) >> 1;
    cdq(le, mid); cdq(mid+1, ri);

    int p = le,q = mid+1;
    int tot = 0;
    while(p <= mid && q <= ri) {
        if(a[p].x <= a[q].x){
            if(a[p].op == 1) {
                update(a[p].x, a[p].y,a[p].val);
                que.push(p);
            }
            tmp[++tot] = a[p++];
        }
        else {
            if(a[q].op == 2) {
                ans[a[q].val] = (ans[a[q].val] + solve(a[q].x, a[q].y) ) % mod;
            }
            else if(a[q].op == 3) {
                ans[a[q].val] = ((ans[a[q].val] - solve(a[q].x, a[q].y) ) ) % mod;
                if(ans[a[q].val]< 0) 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 == 2) {
                ans[a[q].val] = (ans[a[q].val] + solve(a[q].x, a[q].y) ) % mod;
            }
            else if(a[q].op == 3) {
                ans[a[q].val] = ((ans[a[q].val] - solve(a[q].x, a[q].y) ) ) % mod;
                if(ans[a[q].val]< 0) 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, 1, tot) a[i+le-1] = tmp[i];
}
ll ksm(ll a, ll n){
    ll res = 1;
    while(n > 0){
        if(n & 1) res = res * a % mod;
        a = a * a % mod;
        n>>=1;
    }
    return res;
}
int main(){
    int n,m,q;
    scanf("%d%d%d", &n, &m, &q);
    int tot = 0,id = 0;
    while(q--) {
        int op; scanf("%d", &op);
        if(op == 1) {
            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 = 1;
            tot++;  a[tot].x = x1; a[tot].y = y2+1; a[tot].val = -k;a[tot].op = 1;
            tot++;  a[tot].x = x2+1; a[tot].y = y1; a[tot].val = -k;a[tot].op = 1;
            tot++;  a[tot].x = x2+1; a[tot].y = y2+1; a[tot].val = k;a[tot].op = 1;
        }
        else {
            int x1,y1,x2,y2;
            id++;
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            tot++;  a[tot].x = x1-1; a[tot].y = y1-1;   a[tot].val = id; a[tot].op = 2;
            tot++;  a[tot].x = x1-1; a[tot].y = y2; a[tot].val = id; a[tot].op = 3;
            tot++;  a[tot].x = x2; a[tot].y = y1-1; a[tot].val = id; a[tot].op = 3;
            tot++;  a[tot].x = x2; a[tot].y = y2; a[tot].val = id;a[tot].op = 2;
            sz[id] = 1ll*(y2-y1+1)*(x2-x1+1)%mod;
        }
    }

    cdq(1, tot);

    rep(i, 1, id) {
        printf("%lld\n", 1ll*ans[i] * ksm(sz[i], mod-2)%mod);
    }
    return 0;
}