2020杭电多校第八场题解
程序员文章站
2022-03-23 22:37:07
Clockwise or Counterclockwise题目传送门Clockwise or Counterclockwise题目大意给你三个点ABC,判断A->B->C的方向(逆时针或者顺时针)思路叉积/*A(x1,y1) B(x2,y2) C(x3,y3)向量AB=(x2-x1,y2-y1)向量AC=(x3-x1,y3-y1)若 AB x AC < 0 则 ABC 顺时针若 AB x AC > 0 则 ABC 逆时针若 AB x AC = 0 则 AB、...
Clockwise or Counterclockwise
题目传送门
题目大意
给你三个点ABC,判断A->B->C的方向(逆时针或者顺时针)
思路
叉积
/*
A(x1,y1) B(x2,y2) C(x3,y3)
向量AB=(x2-x1,y2-y1)
向量AC=(x3-x1,y3-y1)
若 AB x AC < 0 则 ABC 顺时针
若 AB x AC > 0 则 ABC 逆时针
若 AB x AC = 0 则 AB、AC 共线 同向或者反向
*/
struct node{
double x,y;
}a[N];
double cross(node a,node b,node c){//>0,ab在ac顺时针;<0,ab在ac逆时针
return (b.x-a.x)*(c.y-a.y) - (c.x-a.x)*(b.y-a.y);
}
AC Code
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
#define endl '\n'
#define INF 0x3f3f3f3f
#define int long long
// #define TDS_ACM_LOCAL
const int N=4;
struct node{
double x,y;
}a[N];
double cross(node a,node b,node c){//>0,ab在ac顺时针;<0,ab在ac逆时针
return (b.x-a.x)*(c.y-a.y) - (c.x-a.x)*(b.y-a.y);
}
void solve(){
cin>>a[1].x>>a[1].y>>a[2].x>>a[2].y>>a[3].x>>a[3].y;
double d=cross(a[1], a[2], a[3]);
if(d<0) cout<<"Clockwise"<<endl;
else cout<<"Counterclockwise"<<endl;
return ;
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
#ifdef TDS_ACM_LOCAL
freopen("D:\\VS code\\.vscode\\testall\\in.txt", "r", stdin);
freopen("D:\\VS code\\.vscode\\testall\\out.txt", "w", stdout);
#endif
int T;
cin>>T;
while(T--) solve();
return 0;
}
Fluctuation Limit
题目传送门
题目大意
给你n个区间l和r,以及波动k
求是否有一个序列满足在l~r(可以波动k)
思路
官方题解:
如果 i 是在 [l,r] 范围内, 那么 i + 1 必须要在 [l−k,r + k] 范围内.这是因为如果 i + 1 选了 范围外的值, i 就无解了. 这样可以从左往右, 把左边的约束带到右边.再从右往左做一遍.最后剩下的区间应该就是可 行域.因为题目只要求一种方案, 全部选最低的即可.复杂度 O(n).
AC Code
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
#define endl '\n'
#define INF 0x3f3f3f3f
#define int long long
// #define TDS_ACM_LOCAL
const int N=1e5 +9;
int n, k;
int l[N], r[N], L, R;
void solve(){
cin>>n>>k;
l[0]=0, r[0]=INF;
for(int i=1; i<=n; i++){
cin>>L>>R;
l[i]=max(l[i-1]-k, L);
r[i]=min(r[i-1]+k, R);
}
int flag=1;
for(int i=n-1; i>=1; i--){
l[i]=max(l[i+1]-k, l[i]);
r[i]=min(r[i+1]+k, r[i]);
if(l[i]>r[i]) flag=0;
}
if(flag){
cout<<"YES"<<endl;
for(int i=1; i<=n; i++) cout<<l[i]<<" ";
cout<<endl;
}
else cout<<"NO"<<endl;
return ;
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
#ifdef TDS_ACM_LOCAL
freopen("D:\\VS code\\.vscode\\testall\\in.txt", "r", stdin);
freopen("D:\\VS code\\.vscode\\testall\\out.txt", "w", stdout);
#endif
int T;
cin>>T;
while(T--) solve();
return 0;
}
本文地址:https://blog.csdn.net/xmyrzb/article/details/108130429
推荐阅读
-
【杭电多校2020】1005.Fibonacci Sum(数论,公式)
-
【杭电多校2020】第五场1001.Tetrahedron
-
【杭电多校2020】第二场1001.Total Eclipse(并查集)
-
HDU6759 Leading Robots(2020杭电多校训练第一场)
-
2020杭电多校第五场 1009 Paperfolding
-
2020杭电多校集训-Distinct Sub-palindromes
-
2020牛客多校第八场 K
-
[杭电多校2020]第一场 1004 Distinct Sub-palindromes
-
杭电多校——第二场(题解)
-
2020年杭电多校第五场题解Tetrahedron、Boring Game、Paperfolding