Codeforces Round #427 (Div. 2)C. Star sky

传送门

题意：n个星，q组询问，亮度最大值是c，c+1的时候亮度为0。然后重新亮。下面给你星星的坐标和初始亮度，然后q里面是时间和矩形左下和右上的坐标，问你在这个矩形能看到多少亮度。

思路：二维前缀和，或者存点二分。存点二分还没写，如果有空再写。这里给出思路。存好点之后存入结构体。注：有可能有初始亮度为0的点。

对结构体的点进行排序。x小的y小的在前面。二分符合矩形框的点。然后while处理直到不在矩形范围。

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1e5+10;const int maxx=1e4+100;const double EPS=1e-7;const int MOD=10000007;#define mod(x) ((x)%MOD);template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}inline int Scan(){    int Res=0,ch,Flag=0;    if((ch=getchar())=='-')Flag=1;    else if(ch>='0' && ch<='9')Res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')Res=Res*10+ch-'0';    return Flag ? -Res : Res;}//freopen( "in.txt" , "r" , stdin );//freopen( "data.out" , "w" , stdout );//cerr << "run time is " << clock() << endl;int n,q,c,c1;struct node{    int x,y;}Q[maxn];LL C[22][107][107],s[22][107][107];int t,x1,y3,x2,y2;int main(){    cin>>n>>q>>c;    for(int i=1;i<=n;i++)    {        //cin        int x,y;        x=Scan(),y=Scan();        Q[i].x=x;Q[i].y=y;        c1=Scan();        C[c1][x][y]++;    }    for(int i=0;i<=c;i++)        for(int k=1;k<=105;k++)            for(int z=1;z<=105;z++)                s[i][k][z]=s[i][k][z-1]+                s[i][k-1][z]-s[i][k-1][z-1]+C[i][k][z];    while(q--)    {        t=Scan();x1=Scan();y3=Scan();x2=Scan();y2=Scan();        LL ans=0;        t%=(c+1);        for(int i=0;i<=c;i++)            ans+=((i+t)%(c+1))*(s[i][x2][y2]-s[i][x1-1][y2]                    +s[i][x1-1][y3-1]-s[i][x2][y3-1]);        cout<<ans<<endl;    }}