本文主要是介绍URAL 1707. Hypnotoad's Secret（树状数组），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
URAL 1707. Hypnotoad's Secret
题目链接
题意：这题设置的恶心不能多说，构造点和矩形，大概就是问每个矩形里面是否包含点
思路：树状数组，把点排序，按y轴，在按x轴，在按询问，这样每次遇到一个点就在相应的扫描线上加，遇到查询就询问出左边到这个点位置的，就能预处理出每个点左下角包含的点的个数，然后每个矩形再利用容斥原理去搞一下即可
代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;typedef long long ll;const int N = 1050005;
const int M = 5005;const ll MOD = 200904040930 + 33;struct Point {int x, y;bool isop;Point() {}Point(int x, int y, bool isop) {this->x = x;this->y = y;this->isop = isop;}
} p[N];bool cmp(Point a, Point b) {if (a.y == b.y) {if (a.x == b.x)return a.isop < b.isop;return a.x < b.x;}return a.y < b.y;
}int pn;int n, m;struct OP {int a0, b0, c0, d0;int da, db, dc, dd;int q;void read() {scanf("%d%d%d%d%d%d%d%d%d", &a0, &b0, &c0, &d0, &da, &db, &dc, &dd, &q);}
} op[350];#define lowbit(x) (x&(-x))int bit[M];void add(int x, int v) {while (x <= n) {bit[x] += v;x += lowbit(x);}
}int get(int x) {int ans = 0;while (x) {ans += bit[x];x -= lowbit(x);}return ans;
}int get(int l, int r) {return get(r) - get(l - 1);
}typedef pair<int, int> pii;
#define MP(a,b) make_pair(a,b)map<pii, int> cnt;
int save[350];
ll mi7[350];int main() {mi7[0] = 1;for (int i = 1; i < 350; i++)mi7[i] = mi7[i - 1] * 7 % MOD;while (~scanf("%d%d", &n, &m)) {cnt.clear();pn = 0;memset(bit, 0, sizeof(bit));int s0, t0, ds, dt, k;while (m--) {scanf("%d%d%d%d%d", &s0, &t0, &ds, &dt, &k);for (int i = 0; i < k; i++) {p[pn++] = Point(s0, t0, false);s0 = ((s0 + ds) % n + n) % n;t0 = ((t0 + dt) % n + n) % n;}}scanf("%d", &m);for (int i = 0; i < m; i++) {op[i].read();int a0 = op[i].a0, b0 = op[i].b0, c0 = op[i].c0, d0 = op[i].d0;int da = op[i].da, db = op[i].db, dc = op[i].dc, dd = op[i].dd;int q = op[i].q;for (int j = 0; j < q; j++) {int a = min(a0, b0), b = max(a0, b0), c = min(c0, d0), d = max(c0, d0);p[pn++] = Point(a - 1, c - 1, true);p[pn++] = Point(b, c - 1, true);p[pn++] = Point(a - 1, d, true);p[pn++] = Point(b, d, true);a0 = ((a0 + da) % n + n) % n;b0 = ((b0 + db) % n + n) % n;c0 = ((c0 + dc) % n + n) % n;d0 = ((d0 + dd) % n + n) % n;}}sort(p, p + pn, cmp);for (int i = 0; i < pn; i++) {if (p[i].isop) cnt[MP(p[i].x, p[i].y)] = get(1, p[i].x + 1);else add(p[i].x + 1, 1);}for (int i = 0; i < m; i++) {int a0 = op[i].a0, b0 = op[i].b0, c0 = op[i].c0, d0 = op[i].d0;int da = op[i].da, db = op[i].db, dc = op[i].dc, dd = op[i].dd;int q = op[i].q;for (int j = 0; j < q; j++) {int a = min(a0, b0), b = max(a0, b0), c = min(c0, d0), d = max(c0, d0);int tmp = cnt[MP(b, d)] - cnt[MP(a - 1, d)] - cnt[MP(b, c - 1)] + cnt[MP(a - 1, c - 1)];if (tmp == 0) save[j] = 0;else save[j] = 1;a0 = ((a0 + da) % n + n) % n;b0 = ((b0 + db) % n + n) % n;c0 = ((c0 + dc) % n + n) % n;d0 = ((d0 + dd) % n + n) % n;}if (q <= 20) {for (int j = 0; j < q; j++)printf("%d", save[j]);printf("\n");} else {ll out = 0;for (int j = 0; j < q; j++)out = (out + mi7[j] * save[j]) % MOD;printf("%lld\n", out);}}}return 0;
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