#include <bits/stdc++.h>
using namespace std;
#define re register
const int N = 1e6 + 3;
int a[N];
void solve(){
int n, k;
cin >> n >> k;
for(re int i = 1; i <= n; i ++) cin >> a[i];
int cnt = 0;
for(re int i = 1; i <= k; i ++) if(a[i] > k) cnt ++;
cout << cnt << endl;
}
int main(){
// solve();
int _;
cin >> _;
while(_ --){
solve();
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
#define re register
void solve(){
int n;
cin >> n;
if(n % 2 == 0){
for(re int i = 2; i <= n; i += 2) cout << i << " " << i - 1 << " ";
}
else{
cout << 1 << " ";
for(re int i = 3; i <= n; i += 2) cout << i << " " << i - 1 << " ";
}
cout << endl;
}
int main(){
// solve();
int _;
cin >> _;
while(_ --){
solve();
}
return 0;
}
i
之前的序列一定满足不严格单调递增,在枚举结束之后,b
中元素个数即为操作次数#include <bits/stdc++.h>
using namespace std;
#define re register
const int N = 1e6 + 3;
int a[N];
set<int> b;
void solve(){
int n;
cin >> n;
for(re int i = 1; i <= n; i ++) cin >> a[i];
for(re int i = 2; i <= n; i ++){
if(b.count(a[i]) > 0) a[i] = 0;
if(a[i - 1] > a[i]){
b.insert(a[i - 1]);
a[i - 1] = 0;
for(re int j = i - 1; a[j] != 0 && j >= 1; j --){
b.insert(a[j]);
a[j] = 0;
}
}
}
cout << b.size() << endl;
b.clear();
}
int main(){
// solve();
int _;
cin >> _;
while(_ --){
solve();
}
return 0;
}