Star sky (Codeforces 835C)
Tags: Implement, dp

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.

Output: For each view print the total brightness of the viewed stars.

Note: 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.

Sample Inputs:
2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5

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

Sample Outputs:
3
0
3

3
3
5
0

Solution (Python):
from sys import stdin

def main():
    (n, q, c) = map(int, input().split())
    c += 1
    l = stdin.read().splitlines()
    prec = [[[0] * 101 for _ in range(101)] for _ in range(c)]
    for line in l[:n]:
        (x, y, s) = map(int, line.split())
        for t in range(c):
            prec[t - s][x][y] += t
    del l[:n]
    for plane in prec:
        for x in range(100):
            for y in range(100):
                plane[x + 1][y + 1] += plane[x][y + 1] + plane[x + 1][y] - plane[x][y]
    for (i, line) in enumerate(l):
        (t, x1, y1, x2, y2) = map(int, line.split())
        plane = prec[t % c]
        l[i] = plane[x2][y2] - plane[x1 - 1][y2] - plane[x2][y1 - 1] + plane[x1 - 1][y1 - 1]
    print('\n'.join(map(str, l)))
if __name__ == '__main__':
    main()

Solution (Python):
from sys import stdin

def main():
    (n, q, c) = map(int, input().split())
    c += 1
    l = stdin.read().splitlines()
    prec = [[[0] * 101 for _ in range(101)] for _ in range(c)]
    for line in l[:n]:
        (x, y, s) = map(int, line.split())
        for t in range(c):
            prec[t - s][x][y] += t
    del l[:n]
    for plane in prec:
        for x in range(100):
            for y in range(100):
                plane[x + 1][y + 1] += plane[x][y + 1] + plane[x + 1][y] - plane[x][y]
    for (i, line) in enumerate(l):
        (t, x1, y1, x2, y2) = map(int, line.split())
        plane = prec[t % c]
        l[i] = plane[x2][y2] - plane[x1 - 1][y2] - plane[x2][y1 - 1] + plane[x1 - 1][y1 - 1]
    print('\n'.join(map(str, l)))
if __name__ == '__main__':
    main()

Solution (Python):
from sys import stdin, stdout
import math

def main():
    line = stdin.readline().split()
    n = int(line[0])
    q = int(line[1])
    c = int(line[2])
    dp = [[[0 for i in range(101)] for j in range(101)] for k in range(11)]
    for i in range(n):
        line = stdin.readline().split()
        x = int(line[0])
        y = int(line[1])
        p = int(line[2])
        dp[p][x][y] += 1
    for p in range(c + 1):
        for x in range(1, 101):
            for y in range(1, 101):
                dp[p][x][y] += dp[p][x - 1][y] + dp[p][x][y - 1] - dp[p][x - 1][y - 1]
    for i in range(q):
        line = stdin.readline().split()
        t = int(line[0])
        x1 = int(line[1])
        y1 = int(line[2])
        x2 = int(line[3])
        y2 = int(line[4])
        ans = 0
        for j in range(c + 1):
            bright = (j + t) % (c + 1)
            amount = dp[j][x2][y2] - dp[j][x2][y1 - 1] - dp[j][x1 - 1][y2] + dp[j][x1 - 1][y1 - 1]
            ans += bright * amount
        stdout.write(str(ans) + '\n')
if __name__ == '__main__':
    main()