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

a)
-xi ;
b)
x1 + x2 +...+ xm = b * for some integers a and b (a > 0).
Determine the maximum value of xp1 + xp2 +...+ xpm 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 exists x1, x2,..., xm 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.

#include   
#include   
 
double m, p, a, b, numa, anum, sb, sum; 
int main() { 
    while (~scanf("%lf%lf%lf%lf", &m, &p, &a, &b)) { 
        sum = 0; 
        sb = a * b; 
        numa = anum = 0; 
        for (int i = 0; i < m - 1; i ++) {//注意只进行m - 1次操作，最后一部分要单独考虑  
            if (sb < a) { 
                anum ++; 
                sb ++; 
            } 
            else { 
                sb -= a; 
                numa ++; 
            } 
        } 
        sum += anum / pow(sqrt(a), p); 
        sum += numa * pow(sqrt(a), p); 
        sum += pow((sb / sqrt(a)), p);//剩下的部分  
        printf("%d\n", int(sum + 0.5)); 
    } 
    return 0; 
} 

#include 
#include 

double m, p, a, b, numa, anum, sb, sum;
int main() {
 while (~scanf("%lf%lf%lf%lf", &m, &p, &a, &b)) {
  sum = 0;
  sb = a * b;
  numa = anum = 0;
  for (int i = 0; i < m - 1; i ++) {//注意只进行m - 1次操作，最后一部分要单独考虑
   if (sb < a) {
    anum ++;
    sb ++;
   }
   else {
    sb -= a;
    numa ++;
   }
  }
  sum += anum / pow(sqrt(a), p);
  sum += numa * pow(sqrt(a), p);
  sum += pow((sb / sqrt(a)), p);//剩下的部分
  printf("%d\n", int(sum + 0.5));
 }
 return 0;
}

 

Sample Input

1997 12 3 -318
10 2 4 -1

Sample Output

189548
6题意：给定m,p,a,b.根据题目中的两个条件.求出 xp1 + xp2 +...+ xpm 最大值..

思路:贪心.由于题目明确了p是负数,所以x^p，x绝对值越大的时候值越大。。然后我们根据条件。发现x尽可能取sqrt(a)是最好的。但是不一定能全部取得sqrt(a)。那么多出来的还要拿一部分去抵消。这时候我们就用-1/sqrt(a)去抵消是最好的。这样就能满足最大了。不过要注意。抵消到最后剩下那部分也要考虑进去。

代码：