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这一次我们就简单一点了,题目在此:
2.437
//三分极值 #include<cstdio> #include<algorithm> #include<cmath> //#include<bits/stdc++.h> using namespace std; template<class T>inline T read(T&x) { char c; while((c=getchar())<=32)if(c==EOF)return 0; bool ok=false; if(c=='-')ok=true,c=getchar(); for(x=0; c>32; c=getchar()) x=x*10+c-'0'; if(ok)x=-x; return 1; } template<class T> inline T read_(T&x,T&y) { return read(x)&&read(y); } template<class T> inline T read__(T&x,T&y,T&z) { return read(x)&&read(y)&&read(z); } template<class T> inline void write(T x) { if(x<0)putchar('-'),x=-x; if(x<10)putchar(x+'0'); else write(x/10),putchar(x%10+'0'); } template<class T>inline void writeln(T x) { write(x); putchar('\n'); } //-------ZCC IO template------ const int maxn=1000011; const double inf=999999999; #define lson (rt<<1),L,M #define rson (rt<<1|1),M+1,R #define M ((L+R)>>1) #define For(i,t,n) for(int i=(t);i<(n);i++) typedef long long LL; typedef double DB; typedef pair<int,int> P; #define bug printf("---\n"); #define mod 10007 double px,py; double a,b,c; double d(double x) { return sqrt((x-px)*(x-px)+(a*x*x+b*x+c-py)*(a*x*x+b*x+c-py)); } DB ts(double left,double right) { DB lm,rm,dis; DB y1=0,y2=0; while(left<=right) { dis=(right-left)/3.0; lm=left+dis; rm=lm+dis; y1=d(lm),y2=d(rm); //printf("%lf %lf\n",left,right); //getchar(); if(y1==y2)return y1; else if(y1<y2) { right=rm; } else { left=lm; } } return min(y1,y2); } //0.421 int main() { //#ifndef ONLINE_JUDGE // freopen("in.txt","r",stdin); //#endif // ONLINE_JUDGE int n,m,i,j,t; while(~scanf("%lf%lf%lf%lf%lf",&a,&b,&c,&px,&py)) { if(d(px)==py) printf("%0.00\n"); else printf("%.3lf\n",ts(-200,200)); } return 0; }
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原文地址:http://blog.csdn.net/u013167299/article/details/45033715