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文章目錄

• 題意：
• 思路：

題意：

給你一個數組$a$，你要將其分成四份，讓這四份中和的最大值$?$最小值最小，輸出這個最小值。
$n\le 2e5,a_i\le 1e9$

思路：

直接枚舉不是很好做，考慮先枚舉中間的位置$i$，設被劃分成的四段和分別是$s1,s2,s3,s4$且$s1<s2,s3<s4$，那么答案就是$max(s2,s4)?min(s1,s3)$。由此可知，我們要讓$abs(s2?s1)$和$abs(s4?s3)$都盡可能小，用兩個指針維護一下就好啦。

// Problem: D - Equal Cut // Contest: AtCoder - AtCoder Regular Contest 100 // URL: https://atcoder.jp/contests/arc100/tasks/arc100_b // Memory Limit: 1024 MB // Time Limit: 2000 ms // // Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native") //#pragma GCC optimize(2) #include<cstdio> #include<iostream> #include<string> #include<cstring> #include<map> #include<cmath> #include<cctype> #include<vector> #include<set> #include<queue> #include<algorithm> #include<sstream> #include<ctime> #include<cstdlib> #include<random> #include<cassert> #define X first #define Y second #define L (u<<1) #define R (u<<1|1) #define pb push_back #define mk make_pair #define Mid ((tr[u].l+tr[u].r)>>1) #define Len(u) (tr[u].r-tr[u].l+1) #define random(a,b) ((a)+rand()%((b)-(a)+1)) #define db puts("---") using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); } //void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); } //void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f; const LL inf=0x3f3f3f3f3f3f3f3f; const double eps=1e-6;int n; int a[N]; LL pre[N];LL get(LL a,LL b,LL c,LL d) {return max(a,max(b,max(c,d)))-min(a,min(b,min(c,d))); }int main() { // ios::sync_with_stdio(false); // cin.tie(0);cin>>n;for(int i=1;i<=n;i++) scanf("%d",&a[i]),pre[i]=pre[i-1]+a[i];LL ans=inf;for(int l=0,r=1,i=1;i<=n;i++) {while(l+1<=i&&abs(pre[l]-pre[i]+pre[l])>abs(pre[l+1]-pre[i]+pre[l+1])) l++;while(r+1<=n&&abs(pre[r]-pre[i]-pre[n]+pre[r])>abs(pre[r+1]-pre[i]-pre[n]+pre[r+1])) r++;ans=min(ans,get(pre[l],pre[i]-pre[l],pre[r]-pre[i],pre[n]-pre[r]));}printf("%lld\n",ans);return 0; } /**/

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