学步园

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Description

Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing his toys into the
box. All the toys get mixed up, and it is impossible for Reza to find his favorite toys anymore. 
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top: 


We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.

4 10 0 10 100 0
20 20
80 80
60 60
40 40
5 10
15 10
95 10
25 10
65 10
75 10
35 10
45 10
55 10
85 10
5 6 0 10 60 0
4 3
15 30
3 1
6 8
10 10
2 1
2 8
1 5
5 5
40 10
7 9
0

Box
2: 5
Box
1: 4
2: 1

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题意：

算出每个小格中玩具的数目，数出各种玩具数目（>0）的小格数

分析：

二分查找+叉积

跟上一题基本一样

点，线，面，形基本关系，点积叉积的理解

#include<iostream>

#include<cstring>

#include<cstdio>

#include<algorithm>

using namespace std;

class T

{

public:

    int x1,x2;

};

bool cmp(T a,T b)

{

    return a.x2<=b.x2;

}

T toy[1005];

int g[1005],p[1005];

int y1,y2,n,m,x,y,t;

int f(int x1,int x2)

{

    int f1=(toy[x1].x2-x)*(y1-y)-(y2-y)*(toy[x1].x1-x);

    int f2=(toy[x2].x2-x)*(y1-y)-(y2-y)*(toy[x2].x1-x);

    if(f1<0&&f2>0) return 0;

    else if(f1>0) return -1;//直线左

    else return 1;          //直线右

}

void solve(int left,int right)

{

    int mid=(left+right)/2;

    int k=f(mid,mid+1);

    if(k==0) {t=mid;return ;}

    else if(k<0)solve(left,mid);

    else solve(mid+1,right);

}

int main()

{

    while(1)

    {

        int i<