牛客 545A 小A与最大子段和 & CF 660F Bear and Bowling 4-白红宇

牛客 545A 小A与最大子段和 & CF 660F Bear and Bowling 4

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发布时间：2019-06-14

本文共 1914 字，大约阅读时间需要 6 分钟。

大意: 给定序列$a$, 求选择一个子区间$[l,r]$, 使得$\sum\limits_{i=l}^r(i-l+1)a_i$最大.

$n\le2e5, |a_i|\le 1e7$.

记$s[i]=\sum a[i], m[i]=\sum ia[i]$, $dp[i]$为以$i$为右端点的答案, 有

$\begin{align} \notag dp[i] & =\max\limits_{0\le j<i}\{m[i]-m[j]-j(s[i]-s[j])\} \\ & = m[i]-\min\limits_{0\le j<i}\{m[j]-js[j]+js[i]\} \notag\end{align}$

然后斜率优化.

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                #define REP(i,a,n) for(int i=a;i<=n;++i)#define PER(i,a,n) for(int i=n;i>=a;--i)#define hr putchar(10)#define pb push_back#define lc (o<<1)#define rc (lc|1)#define mid ((l+r)>>1)#define ls lc,l,mid#define rs rc,mid+1,r#define x first#define y second#define io std::ios::sync_with_stdio(false)#define endl '\n'#define DB(a) ({REP(__i,1,n) cout<
                
                 <<' ';hr;})using namespace std;typedef long long ll;typedef pair
                 
                   pii;const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}//headconst int N = 1e6+10;int n, a[N], q[N];ll s[N], m[N], dp[N], g[N];double slope(int i, int j) { return ((double)g[i]-g[j])/(i-j);}int main() { scanf("%d", &n); REP(i,1,n) { scanf("%d", a+i); s[i] = s[i-1]+a[i]; m[i] = m[i-1]+(ll)i*a[i]; g[i] = (ll)i*s[i]-m[i]; } q[++*q] = 0; ll ans = -1e18; REP(i,1,n) { int opt=1,l=2,r=*q; while (l<=r) { if (slope(q[mid],q[mid-1])>=s[i]) opt=mid,l=mid+1; else r=mid-1; } ans = max(ans, m[i]-m[q[opt]]-(ll)q[opt]*(s[i]-s[q[opt]])); while (*q>1&&slope(i,q[*q])>slope(q[*q],q[*q-1])) --*q; q[++*q] = i; } printf("%lld\n", ans);}