QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#309657 | #8022. Walker | Ebiarat# | TL | 0ms | 0kb | C++20 | 2.7kb | 2024-01-20 19:37:45 | 2024-01-20 19:37:45 |
answer
#include <iostream>
#include <string>
#include <set>
#include <map>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <cmath>
#include <queue>
#include <sstream>
#include <ctime>
#include <iterator>
#include <string.h>
#include <stack>
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <fstream>
#include <assert.h>
#include <numeric>
#include <complex>
#include <random>
#include <utility>
#define IOS ios_base::sync_with_stdio(0),cin.tie(0), cout.tie(0);
#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define RFOR(i,a,b) for(int i = (a) - 1; i>=(b);i--)
#define rep(i,n) FOR(i,0,n)
#define PB push_back
#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define VI vector<int>
#define PII pair<int,int>
#define PLL pair<long long,long long>
#define VL vector<long long >
#define FILL(a, value) memset(a, value, sizeof(a))
const int nax = (int)1e3 + 147;
using namespace std;
const int MOD = (int)1e9 + 7;
const int INF = 1e9 +47 ;
const long long LINF = (long long)1e18 + 4747;
typedef long long LL;
vector<int >g[nax];
int ignored[nax];
int ign_cnt[nax];
int sz[nax];
int ans[nax];
map<pair<int,int>,bool> edge;
bool is_vertex[nax];
map<string,int> known;
void dfs(int v)
{
sz[v] = g[v].empty();
ign_cnt[v] = ignored[v];
for(auto to : g[v])
{
dfs(to);
sz[v]+=sz[to];
ign_cnt[v]+=ign_cnt[to];
}
if(ign_cnt[v] == sz[v] && v != 1)ans[v] = 1;
else {
for(auto to : g[v])ans[v] += ans[to];
if(ignored[v])ans[v]++;
}
}
void solve()
{
double n, p1, v1, p2, v2 ;
cin >> n >> p1 >> v1 >> p2 >> v2;
if(p1>p2) {
swap(v1,v2);
swap(p1,p2);
}
double ans1, ans2, ans3, ans5, ans6, ans7;
ans1 = (n + min(p1, n-p1))/v1;
ans2 = (n + min(p2, n-p2))/v2;
ans7 = max((n-p1)/v1, p2/v2 );
double l = p1, r = p2;
while (r-l > 1e-13)
{
double mid1 = l+ (r-l)/3.0;
double mid2 = l+ 2*(r-l)/3.0;
double mid1_ans = (min((mid1-p1), p1) + mid1)/v1;
mid1_ans = max(mid1_ans, (min(n-p2, p2-mid1) + n-mid1)/v2);
double mid2_ans = (min((mid2-p1), p1) + mid2)/v1;
mid2_ans = max(mid2_ans, (min(n-p2, p2-mid2) + n-mid2)/v2);
if (mid1_ans < mid2_ans)
r = mid2;
else
l = mid1;
}
double t_ans = (min((l-p1), p1) + l)/v1;
t_ans = max(t_ans, (min(n-p2, p2-l) + n-l)/v2);
cout << fixed << setprecision(10) << min({ans1, ans2, ans7, t_ans}) << endl;
}
int main() {
IOS;
int tt = 1;
cin >> tt;
while(tt--)
{
solve();
}
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 0
Time Limit Exceeded
input:
2 10000.0 1.0 0.001 9999.0 0.001 4306.063 4079.874 0.607 1033.423 0.847