# 1030 perfect sequence (25 points)

Given a positive integer sequence, and a positive integer p, let the maximum value of the sequence be m, and the minimum value be m. if M ≤ mp, the sequence is said to be perfect.

Now, given the parameter p and some positive integers, please choose as many numbers as possible to form a perfect sequence.

### Input format:

The first line of the input gives two positive integers, N and p, where n (< 10 5) is the number of positive integers and p (< 10 9) is the given parameter. The second line gives n positive integers, each of which is no more than 10 9.

### Output format:

How many numbers can you choose to output in a row can be used to form a perfect sequence.

### Input example:

```10 8
2 3 20 4 5 1 6 7 8 9
```

### Output example:

`8`

To points:

```#include<cstdio>
#include<algorithm>
using namespace std;

typedef long long LL;
const int maxn = 100010;
int A[maxn];

int main(){
LL n, p;
scanf("%lld%lld", &n, &p);
for (int i = 0; i < n; ++i) {
scanf("%d", &A[i]);
}
sort(A, A + n);
int i = 0, j =0, count = 1;
while(j < n && i< n){
while(j < n&& A[i] * p >= A[j]){
count = max(count, j - i + 1);
j++;
}
i++;
}
printf("%d\n", count);
return  0;
}```

Reference code two (dichotomy):

```#include<cstdio>
#include<algorithm>
using namespace std;

typedef long long LL;
const int maxn = 100010;
int A[maxn];
int N, p;

int binartSearch(int i, LL x){
if(A[N - 1] <= x) return  N;
int left = i + 1, right = N -1;
while(left < right){
int mid = (left + right) / 2;
if(A[mid] > x){
right = mid;
}else{
left = mid + 1;
}
}
return  left;
}

int main(){

scanf("%d%d", &N, &p);
for (int i = 0; i < N; ++i) {
scanf("%d", &A[i]);
}
sort(A, A + N);
int ans = 1;

for (int j = 0; j < N; ++j) {
int num = binartSearch(j, (LL)A[j] * p);
ans = max(ans , num - j);
}
printf("%d\n", ans);
return  0;
}```

Reference code 3:

```#include<cstdio>
#include<algorithm>
using namespace std;

typedef long long LL;
const int maxn = 100010;
int A[maxn];
int N, p;

//int binartSearch(int i, LL x){
//    if(A[N - 1] <= x) return  N;
//    int left = i + 1, right = N -1;
//    while(left < right){
//        int mid = (left + right) / 2;
//        if(A[mid] > x){
//            right = mid;
//        }else{
//            left = mid + 1;
//        }
//    }
//    return  left;
//}

int main(){

scanf("%d%d", &N, &p);
for (int i = 0; i < N; ++i) {
scanf("%d", &A[i]);
}
sort(A, A + N);
int ans = 1;

for (int j = 0; j < N; ++j) {
int num = upper_bound(A + j + 1, A + N,(LL)A[j] * p) - A;
ans = max(ans , num - j);
}
printf("%d\n", ans);
return  0;
}```

Tags: Programming

Posted on Wed, 04 Dec 2019 17:25:44 -0500 by Kevmaster