hdu 5116 计数

题目大意：给你n个点， n个点的坐标都在200以内，让你统计不相交的两个L形的种数，且L形的两条边长的gcd = 1。

思路：用二维树状数组维护点的信息，然后划分区块进行统计，题解是用总的减去相交的，不需要用到二维树状数组。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N =  + ;
const int M = 1e4 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +;
const double eps=1e-;
const double pi=acos(-);

int n, m, up[N][N], rt[N][N];
LL cnt[N][N], num[N][N];
bool Map[N][N];

struct BIT  {
    LL a[N][N];
    void modify(int x, int y, LL v) {
        for(int i = x; i < N; i += i & -i)
            for(int j = y; j < N; j += j & -j)
                a[i][j] += v;
    }

    LL sum(int x, int y) {
        LL ans = ;
        for(int i = x; i; i -= i & -i)
            for(int j = y; j; j -= j & -j)
                ans += a[i][j];
        return ans;
    }

    void init() {
        memset(a, , sizeof(a));
    }
}bit1, bit2;

void init() {
    for(int i = ; i < N; i++) {
        for(int j = ; j < N; j++) {
            cnt[i][j] = cnt[i][j - ] + (__gcd(i, j) == );
        }
    }

    for(int i = ; i < N; i++) {
        for(int j = ; j < N; j++) {
            for(int k = ; k <= i; k++) {
                num[i][j] += cnt[k][j];
            }
        }
    }
}

int main() {
    init();
    int T; scanf("%d", &T);
    for(int cas = ; cas <= T; cas++) {
        bit1.init(); bit2.init();
        memset(rt, , sizeof(rt));
        memset(up, , sizeof(up));
        memset(Map, , sizeof(Map));

        scanf("%d", &n);
        for(int i = ; i <= n; i++) {
            int x, y; scanf("%d%d", &x, &y);
            Map[x][y] = true;
        }

        for(int i = ; i >= ; i--) {
            for(int j = ; j >= ; j--) {
                if(Map[i][j]) {
                    up[i][j] = up[i + ][j] + ;
                    rt[i][j] = rt[i][j + ] + ;
                }
            }
        }

        LL ans = ;

        for(int i = ; i >= ; i--) {
            for(int j = ; j <= ; j++) {
                if(!Map[i][j] || up[i][j] <=  || rt[i][j] <= ) continue;
                ans += num[up[i][j] - ][rt[i][j] - ] * (bit2.sum(, ) - bit2.sum(i + up[i][j] - , j));

                int q = i + up[i][j] - ;
                for(int k = i + ; k <=  && k <= q; k++) {
                    ans += cnt[k - i][rt[i][j] - ] * (bit2.sum(q, j) - bit2.sum(k, j));
                    ans += cnt[k - i][rt[i][j] - ] * (bit1.sum(k, j - ));
                }

                bit2.modify(i, j, num[up[i][j] - ][rt[i][j] - ]);
                for(int k = j + ; k <=  && k < j + rt[i][j]; k++) {
                    bit1.modify(i, k, cnt[k - j][up[i][j] - ]);
                }
            }
        }

        printf("Case #%d: %lld\n", cas,  * ans);
    }
    return ;
}

/*
*/