原题链接
Problem Description
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.
Sample 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
Sample Output
3
3
5
0
题目大意
在一个平面上有n颗星星，它们有一个共同的最大亮度，都有自己的坐标和初始亮度，时间每过一秒，亮度就+1，如超过最大亮度就归零。现有q个询问，每个询问给出一个矩形和一个时间，问在这个时间，这个矩形内的星星的亮度总和。
解题思路
考虑到坐标范围比较小，星星的个数和询问数都达到了1e5，所以用一个数组mmp[i][j][p]表示在坐标 （i , j ）上亮度为p的星星的个数。另设一个前缀和数组dp[p][i][j]表示坐标在（1，1）~（i , j ）亮度为p的星星个数和，这个数组可由dp[p][i][j]=dp[p][i-1][j]+dp[p][i][j-1]-dp[p][i-1][j-1]+mmp[i][j][p]这个式子得到。对于每次询问，枚举亮度，通过dp[i][x2][y2]-dp[i][x1-1][y2]-dp[i][x2][y1-1]+dp[i][x1-1][y1-1]找到对应区域内亮度之和，加上时间再取模，再累加即可。
AC代码
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<cmath>
#include<vector>
#include<string>
#include<queue>
#include<list>
#include<stack>
#include<set>
#include<map>
#define ll long long
#define ull unsigned long long
#define db double
//#define rep(i,n) for(int i = 0;i < n; i++)
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define fil(a,b) memset((a),(b),sizeof(a))
#define cl(a) fil(a,0)
#define pb push_back
#define mp make_pair
#define exp 2.7182818
#define PI 3.141592653589793
#define inf 0x3f3f3f3f
#define fi first
#define se second
#define eps 1e-8
#define MOD 1000000007ll
#define sign(x) ((x)>eps?1:((x)<-eps?(-1):(0)))
using namespace std;
int mmp[105][105][12];
int dp[12][105][105];
int main()
{
cl(dp);
int n,q,c;
int tx,ty,tb;
cin>>n>>q>>c;
for(int i=1;i<=n;++i)
{
scanf("%d%d%d",&tx,&ty,&tb);
mmp[tx][ty][tb]++;
}
int time,x1,y1,x2,y2;
for(int i=1;i<=100;++i)
{
for(int j=1;j<=100;++j)
{
for(int p=0;p<=c;++p)
{
dp[p][i][j]=dp[p][i-1][j]+dp[p][i][j-1]-dp[p][i-1][j-1]+mmp[i][j][p];
}
}
}
for(int z=1;z<=q;++z)
{
int res=0;
scanf("%d%d%d%d%d",&time,&x1,&y1,&x2,&y2);
for(int i=0;i<=c;++i)
{
res+=(((i+time)%(c+1))*(dp[i][x2][y2]-dp[i][x1-1][y2]-dp[i][x2][y1-1]+dp[i][x1-1][y1-1]));
}
printf("%d\n",res);
}
return 0;
}