Star sky
The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).
Over time the stars twinkle. At moment 0 the i-th star has brightness si. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.
You want to look at the sky q times. In the i-th time you will look at the moment ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.
A star lies in a rectangle if it lies on its border or lies strictly inside it.
Input
The first line contains three integers n, q, c (1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.
The next n lines contain the stars description. The i-th from these lines contains three integers xi, yi, si (1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.
The next q lines contain the views description. The i-th from these lines contains five integers ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.
OutputFor each view print the total brightness of the viewed stars.
Examples2 3 3 1 1 1 3 2 0 2 1 1 2 2 0 2 1 4 5 5 1 1 5 5
3 0 3
3 4 5 1 1 2 2 3 0 3 3 1 0 1 1 100 100 1 2 2 4 4 2 2 1 4 7 1 50 50 51 51
3 3 5 0
Let's consider the first example.
At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.
At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.
At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.
题意:
给出 n个星星的坐标和它们的亮度si,星星最大亮度为c,t秒后星星的亮度为(si+t)%(c+1),q次询问,每次询问一个矩形区域内t秒后星星的亮度和。
思路:
开一个三维数组,记录每个前缀亮度为si的星星的个数即可
code:
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int sum[105][105][12]; int main(){ int n,q,c; while(cin >> n >> q >> c){ memset(sum,0,sizeof(sum)); for(int i = 1; i <= n; i++){ int x,y,v; cin >> x >> y >> v; sum[x][y][v]++; } for(int i = 1; i <= 100; i++){ for(int j = 1; j <= 100; j++){ for(int k = 0; k <= c; k++){ sum[i][j][k] += sum[i-1][j][k] + sum[i][j-1][k] - sum[i-1][j-1][k]; } } } while(q--){ int t,sx,sy,ex,ey,ans = 0; cin >> t >> sx >> sy >> ex >> ey; for(int i = 0; i <= c; i++){ int x = (i + t) % (c + 1); ans += x * (sum[ex][ey][i] - sum[sx-1][ey][i] - sum[ex][sy-1][i] + sum[sx-1][sy-1][i]); } cout << ans << endl; } } return 0; }