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

/**
* Codeforces
* Problem#835C
* Accepted
* Time:156ms
* Memory:3700k
*/
#include <bits/stdc++.h>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
const signed int inf = (signed)((1u << 31) - 1);
const signed long long llf = (signed long long)((1ull << 61) - 1);
const double eps = 1e-6;
const int binary_limit = 128;
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
template<typename T>
inline boolean readInteger(T& u){
char x;
int aFlag = 1;
while(!isdigit((x = getchar())) && x != '-' && x != -1);
if(x == -1) {
ungetc(x, stdin);
return false;
}
if(x == '-'){
x = getchar();
aFlag = -1;
}
for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0');
ungetc(x, stdin);
u *= aFlag;
return true;
}

int n, q, c;
int xs[100005], ys[100005], ss[100005];
int sum[105][105][11];

inline void init() {
scanf("%d%d%d", &n, &q, &c);
for(int i = 1; i <= n; i++) {
scanf("%d%d%d", xs + i, ys + i, ss + i);
}
}

inline void getPreSum() {
memset(sum, 0, sizeof(sum));
for(int i = 1; i <= n; i++) {
sum[xs[i]][ys[i]][ss[i]]++;
}
for(int i = 1; i <= 100; i++) {
for(int j = 1; j <= 100; j++) {
for(int k = 0; k <= 10; k++)
sum[i][j][k] += sum[i - 1][j][k] + sum[i][j - 1][k] - sum[i - 1][j - 1][k];
}
}
}

inline int getAns(int x, int y, int t0) {
int rt = 0;
for(int i = 0; i <= 10; i++)
rt += (sum[x][y][i]) * ((i + t0) % (c + 1));
return rt;
}

inline void solve() {
int t0, x0, y0, x1, y1;
while(q--) {
scanf("%d%d%d%d%d", &t0, &x0, &y0, &x1, &y1);
printf("%d\n", getAns(x1, y1, t0) - getAns(x0 - 1, y1, t0) - getAns(x1, y0 - 1, t0) + getAns(x0 - 1, y0 - 1, t0));
}
}

int main() {
init();
getPreSum();
solve();
return 0;
}