https://codeforces.com/problemset/problem/1389/D
CD比B水哪里去了
肯定是先选择由一个区间的扩展到覆盖另一个区间,然后再由另一个区间覆盖满整个全区间比较优
那就直接枚举1->n都进行这样的操作,注意可以在一次操作结束后,通过两个区间同时增加的姿势达到k,每次都这样算一下看会不会比继续操作下一对ab区间更优
#include<bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;
const int maxl=3e5+10;
ll n,m,cas,k,cnt,tot;
ll ans,alen,blen,zans;
struct lne
{
ll l,r;
}a,b;
char s[maxl];
bool in[maxl];
inline void prework()
{
scanf("%lld%lld",&n,&k);
scanf("%lld%lld%lld%lld",&a.l,&a.r,&b.l,&b.r);
alen=a.r-a.l+1;blen=b.r-b.l+1;
}
inline void mainwork()
{
ll sum=0;
ll l1=max(a.l,b.l),r1=min(a.r,b.r);
ll l2=min(a.l,b.l),r2=max(a.r,b.r);
ll len=r1-l1,len2=r2-l2;
len=max(0ll,len);
sum=len*n;
ans=0;zans=0;
if(sum>=k)
return;
ll tlen=a.l-l2+b.l-l2+r2-a.r+r2-b.r;
for(int i=1;i<=n;i++)
{
if(sum+(len2-len)>=k)
{
if(l1>r1)
{
if(a.r<b.l)
ans+=b.l-a.r+k-sum;
else
ans+=a.l-b.r+k-sum;
}
else
ans+=k-sum;
if(!zans || ans<zans)
zans=ans;
return;
}
else
{
ans+=tlen,sum+=(len2-len);
if(!zans || ans+(k-sum)*2<zans)
zans=ans+(k-sum)*2;
}
}
if(sum<k)
{
ans+=(k-sum)*2;
if(!zans || ans<zans)
zans=ans;
}
}
inline void print()
{
printf("%lld\n",zans);
}
int main()
{
int t=1;
scanf("%d",&t);
for(cas=1;cas<=t;cas++)
{
prework();
mainwork();
print();
}
return 0;
}