题意:给定一个如上的长方形箱子，中间有n条线段，将其分为n+1个区域，给定m个玩具的坐标，统计每个区域中的玩具个数。
思路:这道题很水，只是要知道会使用叉乘来表示点在线的上面还是下面；
当a.Xmult(b,c) < 0时，表示在线的上面。之后就是二分的时候，不能直接使用mid来ans[mid]++;
因为只是确定点在这条线的两边，到底是哪一边，具体还要用tmp来判断;(模板题)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
using namespace std;
#define MS0(a) memset(a,0,sizeof(a))
const int MAXN = ;
struct point{
int x,y;
point(){}
point(int _x,int _y){
x = _x; y = _y;
}
long long operator *(const point &b)const{// 叉乘
return (1LL*x*b.y - 1LL*y*b.x);
}
point operator -(const point &b)const{
return point(x - b.x,y - b.y);
}
long long dot(const point &b){ //点乘
return 1LL*x*b.x + 1LL*y*b.y;
}
double dist(const point &b){
return sqrt(1LL*(x-b.x)*(x-b.x)+1LL*(y-b.y)*(y-b.y));
}
long long dist2(const point &b){
return 1LL*(x-b.x)*(x-b.x)+1LL*(y-b.y)*(y-b.y);
}
double len(){
return sqrt(1LL*x*x+1LL*y*y);
}
double point_to_segment(point b,point c)//点a到“线段” bc的距离a.point_to_segment(b,c);
{
point v[];
v[] = {c.x - b.x,c.y - b.y};
v[] = {x - b.x,y - b.y};
v[] = {x - c.x,y - c.y};
if(v[].dot(v[]) < ) return v[].len();
if(v[].dot(v[]) > ) return v[].len();
return fabs(.*(v[]*v[])/v[].len());
}
long long Xmult(point b,point c){ // 当a->b与a->c顺时针转时，返回正；
return (b-*this)*(c-*this);
}
void input(){
scanf("%d%d",&x,&y);
}
}p[MAXN]; struct Line{
point s,t;
Line(){}
Line(point _s,point _t){
s = _s,t =_t;
}
}line[MAXN];
int ans[MAXN];
int main()
{
int n,m,i,j,x1,y1,x2,y2,kase = ,U,L;
while(scanf("%d",&n),n){
MS0(ans);
if(kase) puts("");
else kase++;
scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
for(i = ;i <= n;i++){
scanf("%d%d",&U,&L);
line[i] = Line(point(U,y1),point(L,y2));
}
line[] = Line(point(x1,y1),point(x1,y2));
int x,y;
for(i = ;i < m;i++){
scanf("%d%d",&x,&y);
int l = , r = n,mid,tmp;
while(l <= r){
mid = l + r >> ;
if( point(x,y).Xmult(line[mid].s,line[mid].t) < ) r = mid-; //在线的上边
else tmp = mid,l = mid+; //线下的点所在的区域才是改line的标号；
}
ans[tmp]++; }
for(i = ;i <= n;i++){
printf("%d: %d\n",i,ans[i]);
}
}
return ;
}