The beautiful values of the palace(2019南京网络赛)

题目链接：https://nanti.jisuanke.com/t/41298

题意：给一个n * n的螺旋矩阵，n保证是奇数，取一些点使其、获得价值，价值为数位和，然后再给q次查询，求矩阵中的价值总和

题解：树状数组求解，将所有有价值的点和需要处理的有关于矩形的点都先记录下来，关于矩形价值总和的点，一个矩形有四个点。例如x1,y1,x2,y2的价值总和是Sx1,x1 - Sx1- 1,y2 - Sx2,y1 - 1 + Sx2,y2，所以可以有思路先将所有的点按照y的坐标值递增排序，之后就可以保证每次求x轴上的总和就是0，0 到 i，j的价值总和，这样就可以将y处理掉，之后保存答案后输出就好了

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<cstdlib>
#include<queue>
#include<set>
#include<string.h>
#include<vector>
#include<deque>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define eps 1e-4
#define bug printf("*********
")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl;
typedef long long LL;
typedef long long ll;
const int maxn = 1e6 + 5;
const int mod = 998244353;

struct node{
    int flag;
    LL x,y,id;
    LL val;
    friend bool operator < (node a,node b) {
        if(a.y == b.y) {
            if(a.x == b.x) return a.flag < b.flag;
            return a.x < b.x;
        }
        return a.y < b.y;
    }
}p[maxn];

LL re_val(LL x) {
    LL sum = 0;
    while(x > 0) {
        sum += x % 10;
        x /= 10;
    }
    return sum;
}
LL index(LL y,LL x,LL n) {  //求的n * n的矩阵中(x,y)的值是多少
    LL mid = (n + 1) / 2;
    LL p = max(abs(x - mid), abs(y - mid));
    LL ans = n * n - (1 + p) * p * 4;
    LL 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);
    }
}

LL c[maxn],ans[maxn];
void init() {
    memset(c,0,sizeof c);
    memset(ans,0,sizeof ans);
}
LL lowbit(LL x) {
    return x & -x;
}
LL sum(LL i) {
    LL res = 0;
    while(i) {
        res += c[i];
        i -= lowbit(i);
    }
    return res;
}
LL n;

void add(LL i,LL t) {
    while(i <= maxn) {
        c[i] += t;
        i += lowbit(i);
    }
}
int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        init();
        LL m,q;
        scanf("%lld %lld %lld",&n,&m,&q);
        int cnt = 0;
        for(int i = 1;i <= m; i++) {
            LL x,y;
            scanf("%lld %lld",&x,&y);
            p[++cnt] = {0, x, y, -1, re_val(index(x,y,n))};
        }
        for(int i = 1; i <= q; i++) {
            LL x1,y1,x2,y2;
            scanf("%lld %lld %lld %lld",&x1,&y1,&x2,&y2);
            p[++cnt] = {1, x1 - 1, y1 - 1, i, 1};
            p[++cnt] = {1, x1 - 1, y2, i, -1};
            p[++cnt] = {1, x2, y1 - 1, i, -1};
            p[++cnt] = {1, x2, y2, i, 1};
        }
        sort(p + 1, p + 1 + cnt);
        for(int i = 1; i <= cnt; i++) {
            if(p[i].flag == 1) ans[p[i].id] += sum(p[i].x) * p[i].val;
            else add(p[i].x,p[i].val);
        }
        for(int i = 1; i <= q; i++)
            printf("%lld
",ans[i]);
    }
}