[codeforces] 578C Weakness and Poorness || 三分
2017-12-06 19:05
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原题
题目定义了两个变量:
poorness表示一个区间内和的绝对值。
weakness表示一个所有区间最大的poornesss
题目要求你求一个x使得
a1 − x, a2 − x, ..., an − x这个序列的weakness最小
输出最小的weakness
显然,所求值是这样的一个函数
(因为是对abs取max,所以不可能出现有两个谷的不符合要求的函数)
所以我们可以三分,只有当x为ans时,我们才会得到最小值。
已知x时,如何求解最大子段和呢?,因为我们是取abs,所以在记录b[i]为前缀和时,b[i]为结尾的最大子段和要么由前面最小的b[]得到,要么由前面最大的b[]得到。所以我们只要O(n)前缀和然后枚举判断找答案就可以了。
#include<cstdio> #include<algorithm> #include<cmath> #define N 2000010 #define eps 3e-12 using namespace std; int n,a ; double l,r,b ; int read() { int ans=0,fu=1; char j=getchar(); for (;(j<'0' || j>'9') && j!='-';j=getchar()) ; if (j=='-') fu=-1,j=getchar(); for (;j>='0' && j<='9';j=getchar()) ans*=10,ans+=j-'0'; return ans*fu; } double check(double x) { double mn=100000,mx=-100000,ans=0; for (int i=1;i<=n;i++) b[i]=(double)a[i]-x; for (int i=1;i<=n;i++) b[i]+=b[i-1]; ans=fabs(b[1]),mn=min(0.0,b[1]),mx=max(0.0,b[1]); for (int i=2;i<=n;i++) { ans=max(ans,fabs(b[i]-mn)); ans=max(ans,fabs(b[i]-mx)); mn=min(mn,b[i]),mx=max(mx,b[i]); } return ans; } int main() { while (~scanf("%d",&n)) { for (int i=1;i<=n;i++) a[i]=read(); l=-100000; r=100000; while (r-l>eps) { double midl=l+(r-l)/3; double midr=r-(r-l)/3; if (check(midl)<check(midr)) r=midr; else l=midl; } printf("%.9lf\n",check(l)); } return 0; }
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