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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#555932#9253. Prism PalaceJZYZWA 1ms8320kbC++141016b2024-09-10 12:46:242024-09-10 12:46:24

Judging History

你现在查看的是最新测评结果

  • [2024-09-10 12:46:24]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:8320kb
  • [2024-09-10 12:46:24]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
template <typename T>inline void read(T &x)
{
    x=0;char c=getchar();bool f=0;
    for(;c<'0'||c>'9';c=getchar())f|=(c=='-');
    for(;c>='0'&&c<='9';c=getchar())x=(x<<1)+(x<<3)+(c-'0');
    x=(f?-x:x);
}
const int N = 2e5+7;
int n;
typedef double db;
const db pi =acos(-1);
struct node
{
    double x,y;
}p[N],q[N];
node operator -(node a,node b)
{
    return (node){a.x-b.x,a.y-b.y};
}
double arc[N];
double ang(node v)
{
    return atan2(v.y,v.x);
}
double mod(double t)
{
    while(t<0)t+=2.0*pi;
    while(t>2.0*pi)t-=2.0*pi;
    return t;
}
int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
    {
        scanf("%lf %lf",&q[i].x,&q[i].y);
    }
    for(int i=0;i<n;i++)p[i]=q[(i+1)%n]-q[i];
    for(int i=0;i<n;i++)
    arc[i]=mod(pi-ang(p[(i+1)%n])+ang(p[i]));
    db res=0;
    for(int i=0;i<n;i++)
    {
        res+=max(0.0,pi-arc[i]-arc[(i+1)%n]);
    }
    res/=pi;
    printf("%.10lf",res);
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 8320kb

input:

3
0 0
1 0
0 1

output:

1.0000000000

result:

ok found '1.0000000', expected '1.0000000', error '0.0000000'

Test #2:

score: 0
Accepted
time: 1ms
memory: 8292kb

input:

4
0 0
0 1
1 1
1 0

output:

0.0000000000

result:

ok found '0.0000000', expected '0.0000000', error '-0.0000000'

Test #3:

score: -100
Wrong Answer
time: 1ms
memory: 6288kb

input:

4
0 0
0 3
1 2
1 1

output:

0.0000000000

result:

wrong answer 1st numbers differ - expected: '0.5000000', found: '0.0000000', error = '0.5000000'