对于a[i+1]=a[i]+1的问题，通常做法都是b[i]=a[i]-i，这样问题就变成了找连续的相等数组。最小的操作数一定是把它变成该数组的中位数。经典问题了。找中位数可以用树状数组实现，在树状数组上二分即可。计算操作数可以分成两个部分计算，假设区间[l, r]，中位数的下标为mid，中位数的值为val，以中位数为界，分成前后两部分，前面部分为 (mid - l + 1) * val - sum[l, mid]，后面部分为sum[mid+1, r] - (r - mid) * val，其中sum[l, r]表示区间[l, r]的和。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const long long inf=1e18;
const int N=5e5+10;
ll T;
ll n,m,k;
ll ans,a[N];
multiset <ll> s1,s2;
ll zuo,you,zhong;
bool jiou = 0;inline ll read() {ll sum = 0, ff = 1; char c = getchar();while(c<'0' || c>'9') { if(c=='-') ff = -1; c = getchar(); }while(c>='0'&&c<='9') { sum = sum * 10 + c - '0'; c = getchar(); }return sum * ff;
}void chushihua() {s1.clear(); s2.clear();s1.insert(-inf); s2.insert(inf);zuo = you = zhong = 0;jiou = 0;
}inline void add(ll x) {if(!jiou) {ll A = *(--s1.end());ll B = *s2.begin();if(A<=x && x<=B) { zhong = x; }else if(A>x) {s1.erase(s1.find(A));s1.insert(x);zuo += x-A;zhong = A;}else if(B<x) {s2.erase(s2.find(B));s2.insert(x);you += x-B;zhong = B;}}else {if(x>=zhong) {s1.insert(zhong); zuo += zhong;s2.insert(x); you += x;}else {s2.insert(zhong); you += zhong;s1.insert(x); zuo += x;}zhong = 0;}jiou ^= 1;
}inline void del(ll x) {ll A = *(--s1.end());ll B = *s2.begin();if(!jiou) {if(A>=x) {s1.erase(s1.find(x)); zuo -= x;s2.erase(s2.find(B)); you -= B;zhong = B;}else {s2.erase(s2.find(x)); you -= x;s1.erase(s1.find(A)); zuo -= A;zhong = A;}}else {if(zhong==x) {}else if(x>zhong) {s2.erase(s2.find(x));s2.insert(zhong);you += zhong-x;}else {s1.erase(s1.find(x));s1.insert(zhong);zuo += zhong-x;}zhong = 0;}jiou ^= 1;
}int main() {T = read();while(T--) {chushihua();n = read(); k = read();for(ll i=1;i<=n;i++) a[i] = read()-i;a[n+1] = inf;ll now = 1;ans = 1; add(a[1]); jiou = 1;for(ll i=1;i<=n;i++) {while(you-zuo<=k) { ans = max(ans,now-i+1); add(a[++now]); }del(a[i]);}cout<<ans<<endl;}return 0;
}