题解 | 1or0

#include<bits/stdc++.h>
#define int long long
using namespace std;
int const N = 2e5+9;
int const inf = 1e9+7;
string s;
int n,q;
pair<int,int> block[N];
int a[N];
int cnt;


signed main()
{
	cin>>n;
	cin>>s;
	s = " " + s;
	int m = 0,last = -1,l = 1,r = 1;
	for(int i=1;i<=n;i++)
	{
		if(last == s[i] || last == -1)
		{
			last = s[i];
			r = i;
		}
		else 
		{
			if(last == '0')
			{
				block[++cnt] = make_pair(l,r);
			}
			last = s[i];
			l = i,r = i;
		}
	}
	if(last == '0') block[++cnt] = make_pair(l,r);
	for(int i=1;i<=cnt;i++)
	{
		int len = block[i].second - block[i].first + 1;
		a[i] = a[i-1] + (len + 1) * len / 2;
	}
	cin>>q;
	for(int i=1;i<=q;i++)
	{
		int l,r;
		cin>>l>>r;
		if(r < block[1].first || l > block[cnt].second)
		{
			int len = r - l + 1;
			cout<<len * (len + 1) / 2<<endl;
			continue;
		}
		int pl = -1,pr = -1;
		int L = 1,R = cnt;
		while(L < R)
		{
			int m = (L + R + 1) >> 1;
			if(block[m].first > r) R = m - 1;
			else L = m;
		}
		pr = L;
		
		L = 1,R = cnt;
		while(L < R)
		{
			int m = (L + R) >> 1;
			if(block[m].second < l) L = m + 1;
			else R = m;
		}
		pl = R;
		
		int ans = 0;
		if(block[pl].first <= l && l <= block[pl].second)
		{
			int len = block[pl].second - l + 1;
			ans += len * (len + 1) / 2;
			pl++;
		}
		if(block[pr].first <= r && r <= block[pr].second)
		{
			int len = r - block[pr].first + 1;
			ans += len * (len + 1) / 2;
			pr--;
		}
		
		
		if(pl <= pr) ans += a[pr] - a[pl-1];
		int len = r - l + 1;
		cout<<max((len + 1) * len / 2 - ans,0ll)<<endl;
	}
	return 0;
}
/*
1 1
2 2 + 1
3 3 + 2 +1

*/

糖丸了，用二分求了一下左边界向右的第一个0段，右边界向左的第一个0段。

然后处理了下边界情况，即边界在一个0段中。

但是向右，向左两边遍历就可以预处理出i向左，向右的第一个0段。