贪心，尽量让最多的Xi=sqrt(a), 多出来的mod个中mod-1个取-1/sqrt(a)，  mod中的最后一个取和mod-1个-1/sqrt(a)加起来相抵消的正数；
#include <iostream>
#include <cmath>
using namespace std;
_int64 m, p, a, b;

void work()
{
	int bit = a;
	int mus_p=0,mus_n=0;
	if(b>0)
	{
       mus_p = b;
	}
	if(b<0)
	{
		mus_n = -1*b*a;
	}
	int left = m - mus_p-mus_n;
	int pl = left/(a+1);//
	int n = pl*a; //

	int mod1 = left%(a+1);
	int m_n = mod1-1;
	double m_p_t = (double)(m_n)/sqrt((double)a);
    double rst =0.0; 
	rst +=(double)(mus_p+pl) * pow( (double)a,(double)(p/2));
	rst +=(double)(mus_n + n + m_n) * pow(1.0/(double)a,(double)(p/2));
	rst +=pow(m_p_t,(double)p);
	printf("%.0f\n",rst);
}

int main()
{
	while (scanf("%d%d%d%d",&m,&p,&a,&b)!=EOF)
	{
		work();
	}
	return 0;
}

• 因为漏掉了一点.. (919) qzy6 2006-11-05 01:20:20
  • 改了就AC, 谢谢哈 (0) bmexue 2006-11-06 16:17:37
• Re:为什么错？ 我的思路是： (885) LJW 2013-05-08 03:04:07