Little Sub and Triangles
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Little Sub and Triangles
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http://acm.hznu.edu.cn/OJ/problem.php?cid=1263&pid=10
http://acm.hznu.edu.cn/OJ/problem.php?id=2589
題解:暴力求面積*2
二分
/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=5000000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q; int ans,cnt,flag,temp,sum; ll e[N]; ll a[N],b[N]; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){scanf("%lld%lld",&a[i],&b[i]);for(int j=1;j<i;j++){for(int k=j+1;k<i;k++){e[++cnt]=labs((a[i]-a[j])*(b[i]-b[k])-(a[i]-a[k])*(b[i]-b[j]));}}}ll l,r;sort(e+1,e+cnt+1);for(int i=1;i<=m;i++){scanf("%lld%lld",&l,&r);printf("%lld\n",upper_bound(e+1,e+cnt+1,r*2)-lower_bound(e+1,e+cnt+1,l*2));}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }總結
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