【CodeForces】gym-101205B Curvy Little Bottles (2012 ACM-ICPC World Finals B)

最新推荐文章于 2022-01-17 22:09:38 发布

原创最新推荐文章于 2022-01-17 22:09:38 发布 · 487 阅读

0 ·

CC 4.0 BY-SA版权

CodeForces 专栏收录该内容

35 篇文章

订阅专栏

$V=\int\limits_{l}^{r} {\pi f^2(x)dx}$
在f(x)是多项式情况下
$\int\limits_{l}^{r}{f(x)dx}=\sum\limits^{n}_{i=0}\frac{a_i}{i+1}(r^{i+1}-l^{i+1})$
推出以上然后就可以做了，第二部分可以二分做
（代码很丑，是打广工训练的时候“抢”时间写的……）

#include<stdio.h>
#define eps 0.000001

double a[23],b[23],xlow,xhigh,inc,l,r,mid;
int T,n,cs;

double pow(double a,int b)
{
    double rt=1;
    while (b--) rt*=a;
    return rt;
}

inline double V(double x)
{
    double rt=0;
    for (int i=0;i<=2*n;i++) rt+=3.1415926535*b[i]/(i+1)*(pow(x,i+1)-pow(xlow,i+1));
    return rt;
}

inline void prepare()
{
    for (int i=0;i<=20;i++) b[i]=0;
}

int main()
{
    while(~scanf("%d",&n))
    {
        cs++;
        prepare();
        for (int i=0;i<=n;i++) scanf("%lf",a+i);
        for (int i=0;i<=n;i++) for (int j=0;j<=n;j++) b[i+j]+=a[i]*a[j];
        scanf("%lf%lf%lf",&xlow,&xhigh,&inc);
        printf("Case %d: ",cs);
        printf("%.2f\n",V(xhigh));
        double fxxk=xlow;
        for (int i=1;i<=8;i++)
        {
            l=xlow;r=xhigh;
            for (mid=(l+r)/2;l+eps<r;mid=(l+r)/2) if (V(mid)<inc) l=mid;else r=mid;
            if (l+eps<xhigh || inc<eps+V(l)) printf("%.2f ",(xlow=l)-fxxk);
            else
            {
                if (i==1) printf("insufficient volume");
                break;
            }
        }
        puts("");
    }
}