The Preliminary Contest for ICPC China Nanchang National Invitational I : Max answer

https://nanti.jisuanke.com/t/38228

给你一个序列，让你求出一个连续子序列，序列最小值 乘 序列和最大, n5e5, a[i] : [-1e5, 1e5];

只有正数就直接单调栈了，不会看这https://blog.youkuaiyun.com/swunHJ/article/details/89441597

考虑存在负数的情况：

如果a[i]是正的，那就在[l[i] - 1, i-1]这段找个最小的前缀和，在[i, r[i]]这段找个最大的前缀和，相减乘一下维护答案就好了

如果a[i]是负的，那就在[l[i] - 1, i -1]这段找个最大的前缀和，在[i, r[i]]这段找个最小的前缀和，相减乘一下维护答案就好了

没了

转换成了单调栈+RMQ问题，显然可以用线段树，但是直接st做更好写一点

int a[maxn], l[maxn], r[maxn];
ll sum[maxn], stmx[maxn][23], stmn[maxn][23];
int n;

stack<int> st;

void init(){
	rep(i, 0, n) stmx[i][0] = stmn[i][0] = sum[i];
	for(int j = 1; (1 << j) <= n + 1; j++){
		for(int i = 0; i + (1 << j) - 1 <= n; i++){
			stmx[i][j] = max(stmx[i][j-1], stmx[i+(1<<(j-1))][j-1]);
			stmn[i][j] = min(stmn[i][j-1], stmn[i+(1<<(j-1))][j-1]);
		}
	}
	rep(i, 1, n){
		while(!st.empty() && a[i] <= a[st.top()]) st.pop();
		if(st.empty()) l[i] = 1;
		else l[i] = st.top() + 1;
		st.push(i);
	}
	while(!st.empty()) st.pop();
	Rep(i, n, 1){
		while(!st.empty() && a[i] <= a[st.top()]) st.pop();
		if(st.empty()) r[i] = n;
		else r[i] = st.top() - 1;
		st.push(i);
	}
}

ll askmx(int l,int r){
	int k=log2(r-l+1);
	return max(stmx[l][k],stmx[r-(1<<k)+1][k]);
}

ll askmn(int l,int r){
	int k=log2(r-l+1);
	return min(stmn[l][k],stmn[r-(1<<k)+1][k]);
}

int main()
{
	scanf("%d",&n);
	rep(i, 1, n) scanf("%d", a + i), sum[i] = sum[i-1] + a[i];
	init();
	ll ans = -INF;
	rep(i, 1, n){
		if(a[i] < 0){
			ll l2 = askmx(l[i] - 1, i - 1), r2 = askmn(i, r[i]);
			ans = max(ans, 1ll * (r2 - l2) * a[i]);
		} else if(a[i] > 0){
			ll l2 = askmn(l[i] - 1, i - 1), r2 = askmx(i, r[i]);
			ans = max(ans, 1ll * (r2 - l2) * a[i]);
		} else ans = max(ans, 0ll);
	}
	printf("%lld\n", ans);
	return 0;
}