树状数组+二维前缀和（A.The beautiful values of the palace）--The Preliminary Contest for ICPC Asia Nanjing 2019

题意：

给你螺旋型的矩阵，告诉你那几个点有值，问你某一个矩阵区间的和是多少。

思路：

以后记住：二维前缀和sort+树状数组就行了！！！。

#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc  exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\Users\13606\Desktop\草稿.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr
#include <string>
#include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
#define fo(a,b,c) for(register int a=b;a<=c;++a)
#define fr(a,b,c) for(register int a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef long long ll;
void swapp(int &a,int &b);
double fabss(double a);
int maxx(int a,int b);
int minn(int a,int b);
int Del_bit_1(int n);
int lowbit(int n);
int abss(int a);
//const long long INF=(1LL<<60);
const double E=2.718281828;
const double PI=acos(-1.0);
const int inf=(1<<30);
const double ESP=1e-9;
const int mod=(int)1e9+7;
const int N=(int)1e6+10;

ll c[N],ans[N];
void Init(int n)
{
    for(int i=1;i<=n;++i)
        c[i]=ans[i]=0;
}
void add(int i,ll t)
{
    while(i<=N)
    {
        c[i]+=t;
        i+=lowbit(i);
    }
}
ll sum(int i)
{
    ll sum=0;
    while(i)
    {
        sum+=c[i];
        i-=lowbit(i);
    }
    return sum;
}
//-----------------------------------树状数组;
struct node
{
    int f;
    int x,y,ans_id;
    ll val;
    friend bool operator<(node a,node b)
    {
        if(a.y==b.y)
        {
            if(a.x==b.x)
                return a.f<b.f;
            return a.x<b.x;
        }
        return a.y<b.y;
    }
}p[N];
ll re_val(ll x)
{
    ll sum=0;
    while(x>0)
    {
        sum+=x%10;
        x/=10;
    }
    return sum;
}
long long index(long long y,long long x,long long n)
{
    long long mid=(n+1)/2;
    long long p=max(abs(x-mid),abs(y-mid));
    long long ans=n*n-(1+p)*p*4;
    long long sx=mid+p,sy=mid+p;
    if(x==sx&&y==sy)
        return ans;
    else
    {
        if(y==sy||x==sx-2*p)
            return ans+abs(x-sx)+abs(y-sy);
        else
            return ans+8*p-abs(x-sx)-abs(y-sy);
    }
}
void solve(int n)
{
    for(int i=1;i<=n;++i)
    {
        if(p[i].f) ans[p[i].ans_id]+=sum(p[i].x)*p[i].val;
        else add(p[i].x,p[i].val);
    }
}

int main()
{
    int T;
    sc("%d",&T);
    while(T--)
    {
        int n,m,ask;
        sc("%d%d%d",&n,&m,&ask);
        Init(N-3);
        int cnt=0;
        for(int i=1;i<=m;++i)
        {
            int x,y;
            sc("%d%d",&x,&y);
            p[++cnt]={0,x,y,-1,re_val(index(x,y,n))};
        }
        for(int i=1;i<=ask;++i)
        {
            int xl,yl,xr,yr;
            sc("%d%d%d%d",&xl,&yl,&xr,&yr);
            p[++cnt]={1,xl-1,yl-1,i,1};
            p[++cnt]={1,xl-1,yr,i,-1};
            p[++cnt]={1,xr,yl-1,i,-1};
            p[++cnt]={1,xr,yr,i,1};
        }
        sort(p+1,p+1+cnt);
        solve(cnt);
        for(int i=1;i<=ask;++i)
            pr("%lld
",ans[i]);
    }
    return 0;
}

/**************************************************************************************/

int maxx(int a,int b)
{
    return a>b?a:b;
}

void swapp(int &a,int &b)
{
    a^=b^=a^=b;
}

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

int Del_bit_1(int n)
{
    return n&(n-1);
}

int abss(int a)
{
    return a>0?a:-a;
}

double fabss(double a)
{
    return a>0?a:-a;
}

int minn(int a,int b)
{
    return a<b?a:b;
}