Niuke hand speed monthly race 48 C (I can't understand the difference. It belongs to yes)

I don't want to play spicy !
  It's been a long time since I hit a cow , Second question wa5 Hair , The mentality is a little fried . I don't know the penalty time when I do the questions of ladder competition every day , Just a tough man , It's a bad habit . only , Clean up and retire .
subject
The question : Given n A number from 1 To n The range of , Given len, Find how many intervals satisfy the length of len And the number of intersections with the interval is the largest and the weight is the largest .
The weight : Number and of all intersecting sections .
Ideas : Just think about it , Move only one grid at a time , Just preprocess the information . Silly , Must use vector Handle . This is very similar to the difference , But not where it disappears –, Because maybe the left end of the interval is still in the interval . Just open one more array to record the right endpoint of all given intervals , When enumerating, you can subtract .
！ Note that the maximum at the right endpoint is 5e6+5e6
Time complexity : O(n)
Code :

#include<bits/stdc++.h>
using namespace std;
const int N = 1e7+10;
typedef long long ll;
int n,m,k,T;
int b[N],bb[N];
ll w[N],ww[N];
ll ans = 0;
ll mx = 0;
ll quan = 0;
void solve()
{
    
	cin>>n>>m;
	for(int i=1;i<=n;++i)
	{
    
		int x,y; cin>>x>>y;
		b[x]++,bb[y]++;
		w[x]+=i,ww[y]+=i;
	}
	ll now;ll sum;
    for(int i=1;i<=m;++i) mx += b[i],quan += w[i];ans = 1;
    now = mx; sum = quan;
    for(int i=m+1;i<N;++i)
    {
    
    	now += b[i]; sum += w[i];
    	now -= bb[i-m]; sum -= ww[i-m];
    	if(now > mx || (now==mx&&sum>quan))
        {
    mx = now; quan = sum; ans = 0;}
        if(now==mx&&sum==quan) ans++;
    }
	cout<<ans;
}
signed main(void)
{
    
	T = 1;
	// cin>>T;
	while(T--)
	solve();
	return 0;
}