正题

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题目大意

给出$n_0,k,p_0$

然后有两问，求

1. $\lfloor n_0-kp\rfloor (p-p_0)$的最大值，$p$为任意实数。
2. $\lfloor n_0-k{p}_1\rfloor (p_1-p_0)+(\lfloor n_0-k{p}_2\rfloor -n_1)(p_2-p_0)$的最大值其中$n1=\lfloor n_0-k{p}_1\rfloor$。$p_1,p_2$为任意实数。

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解题思路

对于第一问，钦定定义$p^{\prime }=\lfloor kp\rfloor$，然后
$$\lfloor n_0-kp\rfloor (p-p_0)=-k{p}^2+(n_0+k{p}_0)p-n_0{p}_0$$
转换为$p^{\prime }$
$$-\frac{p^{\prime }}{k}+(\frac{n_0}{k}+p_0)p^{\prime }-n_0{p}_0$$
然后$a=-\frac{1}{k},b=\frac{n_0}{k}+p_0,c=-n_0{p}_0$
之后用公式$-\frac{b}{2a}$计算出$p$的最优值。

对于第二问，我们可以暴力枚举$k\ast p_1$然后在用公式计算$p_2$。

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$code$

#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
double n0,p0,k,ans;
int main()
{//freopen("math.in","r",stdin);//freopen("math.out","w",stdout);scanf("%lf%lf%lf",&n0,&p0,&k);double a=-1/k,b=n0/k+p0,c=-p0*n0;double p=-b/(2*a);printf("%.3lf",(n0-round(p))*(round(p)/k-p0));for(double p1=0;p1<=n0;p1++){double n1=(n0-p1),y=(n0-p1)*(p1/k-p0);double a=-1/k,b=p1/k+p0,c=n0*p1/k-n0*p0-p1*p1/k;double p2=-b/(2*a);ans=max((p1-round(p2))*(round(p2)/k-p0)+y,ans);}printf(" %.3lf",ans);
}