【Codeforces 578 C. Weakness and Poorness】三分+最大子段和

最新推荐文章于 2020-02-18 12:40:25 发布

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博客详细介绍了如何解决Codeforces上的题目578C，该题涉及序列的贫乏度和弱度概念。弱度是指序列中所有区间最大贫乏度的值，目标是找到一个x值，使得序列减去x后的弱度最小。博主通过观察得出这是一个谷函数，并采用三分搜索策略配合最大子段和算法来求解，特别地，通过取数组的相反数来处理绝对值情况。在实现过程中，由于精度问题导致超时，最终将精度调整为3e-12后成功通过测试。

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cf578C

• 给定一个序列A
• 一个区间的poorness定义为这个区间内和的绝对值
• weakness等于所有区间最大的poorness
• 求一个x使得，序列A全部减x后weakness最小
• 1 ≤ n ≤ 2 * 1e5

做法首先对样例发现随着x增加答案会增加 x减少答案会减少那么这是一个谷函数可以三分
最大子段和就用最大子段和的算法算但是绝对值只要把数组相反数再求一次就是绝对值的情况了
eps 1e-13 TLE一发 3e-12过的

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
const int MAX_N = 200025;
double arr[MAX_N],b[MAX_N];
int n;
double cal(double x,double arr[])
{
    for(int i = 1;i<=n;++i)
        b[i]= arr[i]-x;
    double tmp = 0,ans = 0;
    for(int i = 1;i<=n;++i)
    {
        if(tmp+b[i]<=0) tmp = 0.0;
        else tmp+=b[i];
        ans = max(ans,tmp);
    }
    for(int i = 1;i<=n;++i)
        b[i]=-b[i];
    tmp = 0.0;
    for(int i = 1;i<=n;++i)
    {
        if(tmp+b[i]<=0) tmp = 0.0;
        else tmp+=b[i];
        ans = max(ans,tmp);
    }
    return ans;
}
double trisection_search(double left,double right)
{
    double midl,midr;
    while(right-left>3e-12)
    {
        midl = (left*2+right)/3.0;
        midr = (left+right*2)/3.0;
        if(cal(midl,arr)<=cal(midr,arr)) right = midr;
        else left = midl;
    }
    return cal(left,arr);
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    scanf("%d",&n);
    for(int i = 1;i<=n;++i) scanf("%lf",&arr[i]);
//    dbg(cal(2,arr));
    printf("%.12f\n",trisection_search(-10001,10001));
//    for(int i = 1;i<=n;++i) dbg2(i,arr[i]);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}