Codeforces Round #427 (Div. 2) Problem C Star sky (Codeforces 835C) - 前缀和

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.

Examples
Input
2 3 3 1 1 1 3 2 0 2 1 1 2 2 0 2 1 4 5 5 1 1 5 5
Output
3 0 3
Input
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
Output
3 3 5 0
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.


  题目大意 天空中有一些星星,每个星星有一个初始亮度,如果一个星星的初始亮度为s, 那么在时刻t, 它的亮度为(s + t) % (c + 1)。有q个询问,询问在某一时刻天空中某个矩形内所有星星的亮度和。

  x, y很小,c很小,而且有趣的是10 * 100 * 100 = 100000,标准cf数据范围。

  所以考虑对每种星星的初始亮度搞一个前缀和。这样对于某一时刻,你可以算出某个矩形内亮度为x的星星数目。

  于是这道题就很简单了。。

  值得高兴的是,终于在考试的时候把C题A掉了。。。好开心。。一直认为C题有毒,每次都会做,每次都挂。

Code

 1 /**
 2  * Codeforces
 3  * Problem#835C
 4  * Accepted
 5  * Time:156ms
 6  * Memory:3700k
 7  */ 
 8 #include <bits/stdc++.h> 
 9 #ifndef WIN32
10 #define Auto "%lld"
11 #else
12 #define Auto "%I64d"
13 #endif
14 using namespace std;
15 typedef bool boolean;
16 const signed int inf = (signed)((1u << 31) - 1);
17 const signed long long llf = (signed long long)((1ull << 61) - 1);
18 const double eps = 1e-6;
19 const int binary_limit = 128;
20 #define smin(a, b) a = min(a, b)
21 #define smax(a, b) a = max(a, b)
22 #define max3(a, b, c) max(a, max(b, c))
23 #define min3(a, b, c) min(a, min(b, c))
24 template<typename T>
25 inline boolean readInteger(T& u){
26     char x;
27     int aFlag = 1;
28     while(!isdigit((x = getchar())) && x != '-' && x != -1);
29     if(x == -1) {
30         ungetc(x, stdin);    
31         return false;
32     }
33     if(x == '-'){
34         x = getchar();
35         aFlag = -1;
36     }
37     for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0');
38     ungetc(x, stdin);
39     u *= aFlag;
40     return true;
41 }
42 
43 int n, q, c;
44 int xs[100005], ys[100005], ss[100005];
45 int sum[105][105][11];
46 
47 inline void init() {
48     scanf("%d%d%d", &n, &q, &c);
49     for(int i = 1; i <= n; i++) {
50         scanf("%d%d%d", xs + i, ys + i, ss + i);
51     }
52 }
53 
54 inline void getPreSum() {
55     memset(sum, 0, sizeof(sum));
56     for(int i = 1; i <= n; i++) {
57         sum[xs[i]][ys[i]][ss[i]]++;
58     }
59     for(int i = 1; i <= 100; i++) {
60         for(int j = 1; j <= 100; j++) {
61             for(int k = 0; k <= 10; k++)
62                 sum[i][j][k] += sum[i - 1][j][k] + sum[i][j - 1][k] - sum[i - 1][j - 1][k];
63         }
64     }
65 }
66 
67 inline int getAns(int x, int y, int t0) {
68     int rt = 0;
69     for(int i = 0; i <= 10; i++)
70         rt += (sum[x][y][i]) * ((i + t0) % (c + 1));
71     return rt;
72 }
73 
74 inline void solve() {
75     int t0, x0, y0, x1, y1;
76     while(q--) {
77         scanf("%d%d%d%d%d", &t0, &x0, &y0, &x1, &y1);
78         printf("%d\n", getAns(x1, y1, t0) - getAns(x0 - 1, y1, t0) - getAns(x1, y0 - 1, t0) + getAns(x0 - 1, y0 - 1, t0));
79     }
80 }
81 
82 int main() {
83     init();
84     getPreSum();
85     solve();
86     return 0;
87 }

 

posted @ 2017-08-01 09:33 阿波罗2003 阅读(...) 评论(...) 编辑 收藏