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！！！。
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\草稿.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr 13 #include <string> 14 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 21 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 22 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 23 #define fo(a,b,c) for(register int a=b;a<=c;++a) 24 #define fr(a,b,c) for(register int a=b;a>=c;--a) 25 #define mem(a,b) memset(a,b,sizeof(a)) 26 #define pr printf 27 #define sc scanf 28 #define ls rt<<1 29 #define rs rt<<1|1 30 typedef long long ll; 31 void swapp(int &a,int &b); 32 double fabss(double a); 33 int maxx(int a,int b); 34 int minn(int a,int b); 35 int Del_bit_1(int n); 36 int lowbit(int n); 37 int abss(int a); 38 //const long long INF=(1LL<<60); 39 const double E=2.718281828; 40 const double PI=acos(-1.0); 41 const int inf=(1<<30); 42 const double ESP=1e-9; 43 const int mod=(int)1e9+7; 44 const int N=(int)1e6+10; 45 46 ll c[N],ans[N]; 47 void Init(int n) 48 { 49 for(int i=1;i<=n;++i) 50 c[i]=ans[i]=0; 51 } 52 void add(int i,ll t) 53 { 54 while(i<=N) 55 { 56 c[i]+=t; 57 i+=lowbit(i); 58 } 59 } 60 ll sum(int i) 61 { 62 ll sum=0; 63 while(i) 64 { 65 sum+=c[i]; 66 i-=lowbit(i); 67 } 68 return sum; 69 } 70 //-----------------------------------树状数组; 71 struct node 72 { 73 int f; 74 int x,y,ans_id; 75 ll val; 76 friend bool operator<(node a,node b) 77 { 78 if(a.y==b.y) 79 { 80 if(a.x==b.x) 81 return a.f<b.f; 82 return a.x<b.x; 83 } 84 return a.y<b.y; 85 } 86 }p[N]; 87 ll re_val(ll x) 88 { 89 ll sum=0; 90 while(x>0) 91 { 92 sum+=x%10; 93 x/=10; 94 } 95 return sum; 96 } 97 long long index(long long y,long long x,long long n) 98 { 99 long long mid=(n+1)/2; 100 long long p=max(abs(x-mid),abs(y-mid)); 101 long long ans=n*n-(1+p)*p*4; 102 long long sx=mid+p,sy=mid+p; 103 if(x==sx&&y==sy) 104 return ans; 105 else 106 { 107 if(y==sy||x==sx-2*p) 108 return ans+abs(x-sx)+abs(y-sy); 109 else 110 return ans+8*p-abs(x-sx)-abs(y-sy); 111 } 112 } 113 void solve(int n) 114 { 115 for(int i=1;i<=n;++i) 116 { 117 if(p[i].f) ans[p[i].ans_id]+=sum(p[i].x)*p[i].val; 118 else add(p[i].x,p[i].val); 119 } 120 } 121 122 int main() 123 { 124 int T; 125 sc("%d",&T); 126 while(T--) 127 { 128 int n,m,ask; 129 sc("%d%d%d",&n,&m,&ask); 130 Init(N-3); 131 int cnt=0; 132 for(int i=1;i<=m;++i) 133 { 134 int x,y; 135 sc("%d%d",&x,&y); 136 p[++cnt]={0,x,y,-1,re_val(index(x,y,n))}; 137 } 138 for(int i=1;i<=ask;++i) 139 { 140 int xl,yl,xr,yr; 141 sc("%d%d%d%d",&xl,&yl,&xr,&yr); 142 p[++cnt]={1,xl-1,yl-1,i,1}; 143 p[++cnt]={1,xl-1,yr,i,-1}; 144 p[++cnt]={1,xr,yl-1,i,-1}; 145 p[++cnt]={1,xr,yr,i,1}; 146 } 147 sort(p+1,p+1+cnt); 148 solve(cnt); 149 for(int i=1;i<=ask;++i) 150 pr("%lld\n",ans[i]); 151 } 152 return 0; 153 } 154 155 /**************************************************************************************/ 156 157 int maxx(int a,int b) 158 { 159 return a>b?a:b; 160 } 161 162 void swapp(int &a,int &b) 163 { 164 a^=b^=a^=b; 165 } 166 167 int lowbit(int n) 168 { 169 return n&(-n); 170 } 171 172 int Del_bit_1(int n) 173 { 174 return n&(n-1); 175 } 176 177 int abss(int a) 178 { 179 return a>0?a:-a; 180 } 181 182 double fabss(double a) 183 { 184 return a>0?a:-a; 185 } 186 187 int minn(int a,int b) 188 { 189 return a<b?a:b; 190 }