如果最后的 mod=0 呢?
如下:
>  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;//       =====here !!!!====if(mod1==0) ?

> 	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;
> }

• 改了就AC, 谢谢哈 (0) bmexue 2006-11-06 16:17:37