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=(<<);
 const double ESP=1e-;
 const int mod=(int)1e9+;
 const int N=(int)1e6+;

 ll c[N],ans[N];
 void Init(int n)
 {
     for(int i=;i<=n;++i)
         c[i]=ans[i]=;
 }
 void add(int i,ll t)
 {
     while(i<=N)
     {
         c[i]+=t;
         i+=lowbit(i);
     }
 }
 ll sum(int i)
 {
     ll sum=;
     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=;
     while(x>)
     {
         sum+=x%;
         x/=;
     }
     return sum;
 }
 long long index(long long y,long long x,long long n)
 {
     long long mid=(n+)/;
     long long p=max(abs(x-mid),abs(y-mid));
     long long ans=n*n-(+p)*p*;
     long long sx=mid+p,sy=mid+p;
     if(x==sx&&y==sy)
         return ans;
     else
     {
         if(y==sy||x==sx-*p)
             return ans+abs(x-sx)+abs(y-sy);
         else
             return ans+*p-abs(x-sx)-abs(y-sy);
     }
 }
 void solve(int n)
 {
     for(int i=;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-);
         int cnt=;
         for(int i=;i<=m;++i)
         {
             int x,y;
             sc("%d%d",&x,&y);
             p[++cnt]={,x,y,-,re_val(index(x,y,n))};
         }
         for(int i=;i<=ask;++i)
         {
             int xl,yl,xr,yr;
             sc("%d%d%d%d",&xl,&yl,&xr,&yr);
             p[++cnt]={,xl-,yl-,i,};
             p[++cnt]={,xl-,yr,i,-};
             p[++cnt]={,xr,yl-,i,-};
             p[++cnt]={,xr,yr,i,};
         }
         sort(p+,p++cnt);
         solve(cnt);
         for(int i=;i<=ask;++i)
             pr("%lld\n",ans[i]);
     }
     return ;
 }

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

 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-);
 }

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

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

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