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Time Limit: 20 Sec
Memory Limit: 256 MB
http://codeforces.com/gym/100187/problem/M
Heaviside function is defined as the piecewise constant function whose value is zero for negative argument and one for non-negative argument:
You are given the function f(x) = θ(s1x - a1) + θ(s2x - a2) + ... + θ(snx - an), where si = ± 1. Calculate its values for argument values x1, x2, ..., xm.
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of the summands in the function.
Each of the next n lines contains two integers separated by space — si and ai (si = ± 1, - 109 ≤ ai ≤ 109) — parameters of the i-th summand.
The next line contains a single integer m (1 ≤ m ≤ 200000) — the number of the argument values you should calculate the value of the function for.
The last line contains m integers x1, ..., xm ( - 109 ≤ xi ≤ 109) separated by spaces — the argument values themselves.
Output
Output m lines. i-th line should contain the value of f(xi).
Sample Input
6
1 3
-1 2
1 9
-1 2
1 7
-1 2
8
0 12 2 8 4 -3 7 9
Sample Output
0
3
0
2
1
3
2
3
题意
if(sx-a>=0)ans++;问你每个数通过这个公式，最后的ans是多少
题解：
当s=1时，很显然这个直线是单调向上的，我们可以利用two pointer来优化就好了
复杂度O(n+m)
代码
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 200101 #define mod 1000000009 #define eps 1e-9 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } //************************************************************************************** struct node { int x,y; }; bool cmp1(node a,node b) { return a.x<b.x; } bool cmp2(node a,node b) { return a.y<b.y; } node a[maxn]; vector<int> a1; vector<int> a2; node b[maxn]; int ans[maxn]; int main() { int flag1=0,flag2=0; int n=read(); for(int i=0;i<n;i++) { a[i].x=read(),a[i].y=read(); if(a[i].x==1) flag1++,a1.push_back(a[i].y); else flag2++,a2.push_back(a[i].y); } int m=read(); for(int i=0;i<m;i++) b[i].x=read(),b[i].y=i; sort(b,b+m,cmp1); sort(a1.begin(),a1.end()); sort(a2.begin(),a2.end()); for(int i=m-1;i>=0;i--) { if(flag1==0) break; while(b[i].x<a1[flag1-1]&&flag1>0) flag1--; if(flag1==0) break; ans[b[i].y]+=flag1; } for(int i=0;i<m;i++) { if(flag2==0) break; while(b[i].x>-a2[flag2-1]&&flag2>0) flag2--; if(flag2==0) break; ans[b[i].y]+=flag2; } for(int i=0;i<m;i++) printf("%d\n",ans[i]); }
Codeforces Gym 100187M M. Heaviside Function two pointer
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原文地址：http://www.cnblogs.com/qscqesze/p/4657746.html
