Codeforces Round #320 (Div. 1) [Bayan Thanks-Round] C A Weakness and Poorness

標籤：

f(x)是個凹函數，三分即可，計算方案的時候dp一下。掛精度很坑，指定迴圈次數正解？

#include<bits/stdc++.h>using namespace std;const double eps = 1e-11;const int maxn = 2e5+5;double a[maxn];double d1[maxn],d2[maxn];int n;inline double cal(double x){    double a1 = 0., a2 = 0.;    for(int i = 1; i <= n; i++){        d1[i] = max(0.,d1[i-1])+a[i]-x;        a1 = max(a1,d1[i]);        d2[i] = min(0.,d2[i-1])+a[i]-x;        a2 = min(a2,d2[i]);    }    return max(fabs(a1),fabs(a2));}int main(){    //freopen("in.txt","r",stdin);    scanf("%d",&n);    for(int i = 1; i <= n; i++) scanf("%lf",a+i);    double L = -1e4, R = 1e4;    double M1v,M2v;    while(R-L>eps){        double M1 = L+(R-L)/3, M2 = L+(R-L)/3*2;        M1v = cal(M1), M2v = cal(M2);        if(abs(M1v-M2v)<eps){            L = M1; R = M2;        }else {            if(M1v > M2v){                L = M1;            }else {                R = M2;            }        }    }    printf("%.15lf",(cal(L)+cal(R))/2);    return 0;}

 

Codeforces Round #320 (Div. 1) [Bayan Thanks-Round] C A Weakness and Poorness