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QOJ
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 |
---|---|---|---|---|---|---|---|---|---|
#18428 | #2267. Jewelry Size | guobo# | WA | 9ms | 3720kb | C++ | 1.3kb | 2022-01-19 17:38:08 | 2022-05-06 01:04:34 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> PII;
const int maxn=111111,mod=998244353;
const ld PI=3.14159265358979323846;
#define MP make_pair
#define PB push_back
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define ROF(i,a,b) for(int i=(a);i>=(b);i--)
#define MEM(x,v) memset(x,v,sizeof(x))
inline ll read(){
char ch=getchar();ll x=0,f=0;
while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
return f?-x:x;
}
inline int qpow(int a,int b){
int ans=1;
for(;b;b>>=1,a=1ll*a*a%mod) if(b&1) ans=1ll*ans*a%mod;
return ans;
}
int n;
ld a[maxn];
inline ld calc(ld x,int tp){
ld s=0;
FOR(i,1,n){
ld y=1-a[i]*a[i]/2/x/x,z=acos(y);
s+=z*(i==n?tp:1);
}
return s;
}
int main(){
// n=read();
// FOR(i,1,n) a[i]=read();
// sort(a+1,a+n+1);
n=1000;
FOR(i,1,n-1) a[i]=6;
a[n]=5993;
ld l=a[n]/2,r=1e9;
if(calc(l,1)>=2*PI){
FOR(_,1,100){
ld mid=(l+r)/2;
if(calc(mid,1)<2*PI) r=mid;
else l=mid;
}
printf("%.8lf\n",(double)l);
}
else{
FOR(_,1,100){
ld mid=(l+r)/2;
if(calc(mid,-1)<0) l=mid;
else r=mid;
}
printf("%.8lf\n",(double)l);
}
}
詳細信息
Test #1:
score: 0
Wrong Answer
time: 9ms
memory: 3720kb
input:
5 3 1 6 1 7
output:
94723.64340422
result:
wrong answer 1st numbers differ - expected: '3.5444044', found: '94723.6434042', error = '26723.8411936'