Maximum 贪心

Maximum
Time
Limit:3000MS     Memory
Limit:0KB     64bit IO
Format:%lld & %llu

Submit Status

Description

Let x₁, x₂,..., x_m be
real numbers satisfying the following conditions:

a)

-x_i ;

b)

x₁ + x₂ +...+ x_m = b *  for some integers a and b (a >
0).

Determine the maximum value of x^p₁ + x^p₂ +...+ x^p_m for
some even positive integer p.

Input

Each input line contains four integers: m, p, a, b ( m2000, p12, p is even).
Input is correct, i.e. for each input numbers there existsx₁, x₂,..., x_m satisfying
the given conditions.

Output

For each input line print one number - the maximum value of expression, given
above. The answer must be rounded to the nearest integer.

Sample Input

1997 12 3 -318

10 2 4 -1

Sample Output

189548

6

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<set>

 4 #include<math.h>

 5 #include<iostream>

 6 #include<algorithm>

 7 using namespace std;

 8 int main()

 9 {

10     int m,p,a,b;

11     double ans;

12     while(~scanf("%d%d%d%d",&m,&p,&a,&b))

13     {

14         if(b<0)

15             b=-b,ans=a*b*pow(sqrt(a*1.0)/a,p),m-=a*b;

16         else ans=b*pow(sqrt(a*1.0),p),m-=b;

17         int rr=m/(a+1);

18         m%=(a+1);

19         m--;

20         ans+=rr*(pow(sqrt(a*1.0),p*1.0)+a*pow(sqrt(a*1.0)/a,p*1.0));

21         if(m>0){

22         ans+=m*pow(sqrt(a*1.0)/a,p);

23         ans+=pow(m*sqrt(a*1.0)/a,p);

24         }

25         printf("%.0lf\n",ans);

26     }

27 }

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