1002:
Title:
Given n n intervals, we need to go through these intervals one by one. Notice that they are in turn. I didn't see them at first, and I ignored them.
Explanation:
Find the leftmost and rightmost position of each movement, and judge the relationship between the area to be reached and the current left-right position.
Examples:
The first left is 0, the right is 1000000, ans=0, because it can be placed at will in this interval.
Adding the first interval [1,10], we find that this interval intersects the current interval, and the ans does not need to update the operation, at this time only need to update the interval to [1,10].
Add a second interval [20,30], no intersection with the current interval, update the ans, and update the interval to [20,20]
Why update to [20,20]? Because you have completed the minimum steps and the final coordinate of the class in the target interval can only be 20. Of course, if the final interval is [19,30], the new interval you updated is [19,20].
Here's the code.

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=1001;
struct road{
    int l,r;
    bool operator<(const road& a)const {
        if(l!=a.l)return l<a.l;
        return r<a.r;
    }
}r[maxn];
int ans;
int n;
void dfs(int i,int leftpos,int rightpos){
    if(i==n+1)return;
    if(r[i].l<=leftpos&&r[i].r>=leftpos||r[i].l<=rightpos&&r[i].r>=rightpos||r[i].l>=leftpos&&r[i].l<=rightpos||r[i].r>=leftpos&&r[i].r<=rightpos){
        leftpos=max(leftpos,r[i].l);
        rightpos=min(rightpos,r[i].r);
        dfs(i+1,leftpos,rightpos);
    }
    else{
        int lastpos=r[i].l>rightpos?r[i].l:r[i].r;
        int len=min(abs(r[i].l-rightpos),abs(r[i].r-leftpos));
        ans+=len-len/2;
        if(len%2==0)dfs(i+1,lastpos,lastpos);
        else{
            if(r[i].l!=r[i].r){
                if(r[i].l>rightpos)dfs(i+1,lastpos,lastpos+1);
                else dfs(i+1,lastpos-1,lastpos);
            }
            else dfs(i+1,lastpos,lastpos);
        }
    }
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int a,b;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d%d",&r[i].l,&r[i].r);
        }
        ans=0;
        int leftpos=0,rightpos=1000000;
        dfs(1,leftpos,rightpos);
        printf("%d\n",ans);
    }
    return 0;
}

Just make a list of five questions for one rule-finding question.
The law is one formula for every six numbers.

		scanf("%lld",&n);
        ll x=ceil(1.0*n/6);
        n%=6;
        if(n==0){
            printf("%lld\n",3*x);
        }
        else if(n==1){
            printf("%lld\n",4*x-3);
        }
        else if(n==2){
            printf("%lld\n",3*x-2);
        }
        else if(n==3){
            printf("%lld\n",x-1);
        }
        else if(n==4){
            printf("%lld\n",6*x-3);
        }
        else if(n==5){
            printf("%lld\n",x-1);
        }