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.
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.
For each view print the total brightness of the viewed stars.
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 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.
題意：給出二維面上星星的亮度，每次詢問t秒後，某區域內星星亮度總和，其中亮度變化s=(s+t)%(c+1)
思路：一個點上可能有多個點，定義一個數組d[c][x][y]，表示（0,0）->（x,y）區域上亮度爲c的總個數，狀態轉移方程爲：dp[k][i][j] += dp[k][i - 1][j] + dp[k][i][j - 1] - dp[k][i - 1][j - 1]。利用這些前綴和相減，就能使每次詢問得出結果複雜程度降到O（1）複雜度
上代碼
#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define maxn 105
int dp[12][maxn][maxn];
int main()
{
//freopen("Text.txt","r",stdin);
int n, q, c,x,y,s,x1,y1,t;
memset(dp, 0, sizeof(dp));
scanf("%d%d%d", &n, &q, &c);
while (n--)
{
scanf("%d%d%d", &x, &y, &s);
dp[s][x][y]++;
}
for (int i = 1; i <= 100; i++)
{
for (int j = 1; j <= 100; j++)
{
for (int k = 0; k <= c; k++)
{
dp[k][i][j] += dp[k][i - 1][j] + dp[k][i][j - 1] - dp[k][i - 1][j - 1];//是指（0,0）到（x,y）區域內亮度爲k的總個數
//printf("%d ", dp[k][i][j]);
}
//printf("\n");
}
//printf("\n");
}
while (q--)
{
int sum[12],cnt=0;
scanf("%d%d%d%d%d", &t, &x, &y, &x1, &y1);
for (int i = 0; i <= c; i++)
{
sum[i] = dp[i][x1][y1] + dp[i][x - 1][y - 1] - dp[i][x1][y - 1] - dp[i][x - 1][y1];//畫圖好理解些，因爲邊緣點要包括
cnt += ((i + t)%(c+1))*sum[i];
}
printf("%d\n", cnt);
}
return 0;
}