Codeforces Round #311 (Div. 2)

Water A-ilya and diplomas

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream>using namespace Std;int A[4];int Main (void)  {    int n;  CIN >> N;    int Mn1, MX1, Mn2, MX2, Mn3, mx3;    CIN >> mn1 >> mx1;    CIN >> Mn2 >> MX2;    CIN >> mn3 >> mx3;    A[3] = Mn3; A[2] = Mn2;    int res = n-a[2]-a[3];    if (res <= mx1) a[1] = res;    else    {        a[1] = mx1; Res-= a[1];        if (A[2] + res <= MX2)  a[2] + = res;        else    {            a[2] = MX2; A[3] + = (RES-(a[2]-Mn2));}    }    cout << a[1] << "<< a[2] <<" "<< a[3] << Endl;    return 0;}

Greed | | Dichotomy B-pasha and Tea

Test instructions: There are n girl and N boy tea, cup capacity, boy drink is girl twice times and boy drink as much, girl drink as much, ask Master can pour out how much water

Analysis: The first reaction is to use two-point search girl drink tea capacity, unfortunately write rubbing, mid should and the minimum amount of tea girl and boy to compare. Read the code 99% is greedy, dizzy ~

Code (two points):

#include <cstdio> #include <algorithm> #include <cstring> #include <iostream>using namespace std;const int N = 1e5 + 10;const int INF = 0x3f3f3f3f;const double EPS = 1e-9;int a[n*2];int main (void)  {    int n;
   int W;   scanf ("%d%d", &n, &w);    n *= 2;    for (int i=1; i<=n; ++i)  {        scanf ("%d", &a[i]);    }    Sort (a+1, a+1+n);    Double L = 0.0, r = w;    Double mid = 0.0, ans = 0.0;    for (int i=1; i<=100; ++i)   {        mid = (L + r)/2.0;        If (Mid >= 0 && mid <= a[1] && mid * 2 <= a[n/2+1] && mid * 1.5 * n <= W)    {            L = Mid;            Ans = Mid * 1.5 * n;        }        else    r = Mid;    }    printf ("%.6f\n", ans);    return 0;}

Code (GREEDY):

#include <cstdio> #include <algorithm> #include <cstring> #include <iostream>using namespace std;const int n = 1e5 + 10;int a[n*2];int main (void)  {    int n, W;   scanf ("%d%d", &n, &w);    n *= 2;    for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);    Sort (a+1, a+1+n);    Double ans = min (a[1] * 1.0, A[N/2+1]/2.0);    ans = min (ans * 3 * n/2.0, W * 1.0);    printf ("%.6f\n", ans);    return 0;}

Not to be continued ~

Codeforces Round #311 (Div. 2)