Codeforces 835C Star sky【思维+暴力预处理二维前缀和】

C. Star sky

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 31 1 13 2 02 1 1 2 20 2 1 4 55 1 1 5 5

output

303

input

3 4 51 1 22 3 03 3 10 1 1 100 1001 2 2 4 42 2 1 4 71 50 50 51 51

output

3350

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：

给你N个点，每个点的坐标已知 ，每个点的初始亮度已知。

每过一单位时间，点的亮度都会从si变到si+1.如果si+1>c，那么亮度会变成0..

现在有Q个询问，表示询问区间内，过了ti单位时间之后，亮度的和。

思路：

观察到C的值不大，那么我们设定a【c】【x】【y】表示在点（x，y）处 ，初始亮度为C的点的个数。

那么我们维护C个二维前缀和出来，然后我们每次查询分点初始亮度作为种类去查询即可。

初始亮度为Si的点，过了ti之后贡献的亮度就是(Si+ti)% (c+1)；

那么我们分C种去查询就行，时间复杂度O（c*100*100）；

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;#define ll __int64ll a[15][150][150];ll sum[15][150][150];ll n,q,c;void Slove(){    memset(sum,0,sizeof(sum));    for(ll z=0;z<=c;z++)    {        for(ll i=1;i<=100;i++)        {            for(ll j=1;j<=100;j++)            {                sum[z][i][j]+=sum[z][i][j-1]+a[z][i][j];            }        }    }    for(ll z=0;z<=c;z++)    {        for(ll i=1;i<=100;i++)        {            for(ll j=1;j<=100;j++)            {                sum[z][j][i]+=sum[z][j-1][i];            }        }    }}ll Get(ll z,ll x,ll y,ll xx,ll yy){    ll ans=sum[z][xx][yy]+sum[z][x-1][y-1]-sum[z][x-1][yy]-sum[z][xx][y-1];    return ans;}int main(){    while(~scanf("%I64d%I64d%I64d",&n,&q,&c))    {        memset(a,0,sizeof(a));        for(ll i=1;i<=n;i++)        {            ll x,y,w;            scanf("%I64d%I64d%I64d",&x,&y,&w);            a[w][x][y]++;        }        Slove();        while(q--)        {            ll ti,x,y,xx,yy;            scanf("%I64d%I64d%I64d%I64d%I64d",&ti,&x,&y,&xx,&yy);            ll output=0;            for(ll i=0;i<=c;i++)            {                ll val=(ti+i)%(c+1);                ll summ=Get(i,x,y,xx,yy);                output+=summ*val;            }            printf("%I64d\n",output);        }    }}