LA7147 World Cup Math

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Test instructions: Given sample number T

Number of teams given N, number of qualifying teams m

Given A,b,c (for winning, draw, or losing points respectively)

The maximum number of points that can be obtained, and the smallest possible score to qualify for promotion, respectively.

Idea: Consider the maximum possible score that cannot be promoted, (1) B>=max (a,c) so max= (n-1) *b; (2) B<max (A,C), then first we have to let the former (m+1) team dump

Open the back of the team, that is, Max was initially (N-M-1) *max (a,c), and then the former (m+1) team to carry out M-game, in order to get the most points can not be promoted, then this (m+1) detachment

To the extent possible, that is, to win a losing one, set the current stage of a team to win the X-field, then for each troop in this stage can be divided into ax+ (m-2x) b+cx,0<=x<=m/2;

If a+c>=2b, x take M/2; When a+c<2b, X takes 0. When M is an odd number, it is necessary to calculate the fraction of this stage before the special treatment, namely Max+=max (B,min (A,C)), which is

For this reason can not be guaranteed to win a time to lose one, so for can not qualify through a defeat can lead to not qualify, to take min (a,c), but can be a draw (that is, m+1

The detachment is the same as the section M score), so the summation of the special sentence is Max (B,min (a,c)). Can be promoted to obtain the smallest score can also be similar to the introduction, after consideration (N-M+1) detachment

(B,max (A,c)). See the code:

/********************************************************* file Name:LA7147.cpp Author:kereo create time:2015 year 0 April 30 Thursday 16:41 22 sec *********************************************************/#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <set> #include <map> #include < vector> #include <stack> #include <cmath> #include <string> #include <algorithm> using
namespace Std;
typedef long Long LL;
const int sigma_size=26;
const int N=100+50;
const int MAXN=100000+50;
const int INF=0X3FFFFFFF;
const double eps=1e-8;
const int mod=1000000000+7;
#define L (x) (x<<1) #define R (x) (x<<1|1) #define PII pair<int, int> #define MK (x, y) Make_pair (())
ll N,m,a,b,c;
	int main () {//freopen ("In.txt", "R", stdin);
	int t,kase=0;
	scanf ("%d", &t);
		while (t--) {scanf ("%lld%lld", &n,&m);
		scanf ("%lld%lld%lld", &a,&b,&c);
		ll Ans1,ans2; if (B>=max (a,c)) ans1= (n1) *b;
			else{ans1= (n-m-1) *max (A,C);
			int k=m;
				if (k%2 = = 1) {k--;
			Ans1+=max (B,min (a,c));
			} if (a+c>=2*b) ans1+= (a+c-2*b) *k/2+b*k;
		else Ans1+=b*k;
		} if (B<=min (a,c)) ans2= (n-1) *b;
			else{ans2= (m-1) *min (A,C);
			int k=n-m;
				if (k%2 = = 1) {k--;
			Ans2+=min (B,max (a,c));
			} if (a+c>=2*b) ans2+=b*k;

		else ans2+= (a+c-2*b) *k/2+b*k;

	} printf ("Case #%d:%lld%lld\n", ++kase,ans1,ans2);
} return 0; }