Codeforces Round #364 (Div. 2), problem: (D) As Fast As Possible
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题意:
一群人去一个地方,只有一辆车,已知人的速度,路程,车的速度,人数和每一次车的最大载客量,且每个人只能坐一次车。求最短到达时间。
题解:
人的行程分为走路和坐车,当且仅当走路的人和最后一批坐车的人同时到达时,时间最短,按照对称性,可以知道每一批人坐车和走路的时间都是一样的,所以最后是推公式得到坐车和走路的时间。
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>#include <string>#include <string.h>using namespace std;typedef pair<int, int> p;typedef long long ll;typedef unsigned long long ull;#define pr(x) cout << #x << ": " << x << " "#define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a))#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++)int main(){ int n,k; double l,v1,v2; cin>>n>>l>>v1>>v2>>k; int step; if(n%k==0){step = n/k;} else{step = n/k+1;} double a=(v1+v2)*l/(v1+v2+2*(step-1)*v1); double ans = a/v2+(l-a)/v1; printf("%.10f\n",ans);}
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