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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
#309657#8022. WalkerEbiarat#TL 0ms0kbC++202.7kb2024-01-20 19:37:452024-01-20 19:37:45

Judging History

你现在查看的是最新测评结果

  • [2024-01-20 19:37:45]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:0kb
  • [2024-01-20 19:37:45]
  • 提交

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();
    }
}

詳細信息

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

output:


result: