Hdu-5806-nanoapelovesequence Ⅱ (ruler method)

NanoApe Loves Sequence Ⅱ

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 1348    Accepted Submission(s): 588

Problem Description

NanoApe, the Retired Dog, has returned back to prepare for for the National Higher Education Entrance Examination!

In math class, NanoApe picked up sequences once again. He wrote down a sequence with  n numbers and a number  m on the paper.

Now he wants to know the number of continous subsequences of the sequence in such a manner that the  k-th largest number in the subsequence is no less than  m.

Note : The length of the subsequence must be no less than  k.

 

Input

The first line of the input contains an integer  T, denoting the number of test cases.

In each test case, the first line of the input contains three integers  n,m,k.

The second line of the input contains  n integers  A1,A2,...,An, denoting the elements of the sequence.

1≤T≤10, 2≤n≤200000, 1≤k≤n/2, 1≤m,Ai≤109

 

Output

For each test case, print a line with one integer, denoting the answer.

 

Sample Input

  
   

    
   1
7 4 2
4 2 7 7 6 5 1
  
   

 

Sample Output

  
   

    
   18
  
   

 

Source

BestCoder Round #86

Ruler method :
　　 The whole process is divided into 4 Step ：
　　　　1. Initialize left and right endpoints
　　　　2. Expanding the right endpoint , Until the conditions are met
　　　　3. If the conditions cannot be met in the second step , Then terminate , Otherwise, update the results
　　　　4. Expand the left endpoint 1, Then go back to step two

#include <stdio.h>
int a[200010];
int main(){
    int t,i;
    int n,m,k;
    int l,r,tmp;
    long long sum;

    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&m,&k);
        for(i=0;i<n;i++){
            scanf("%d",a+i);
        }
        sum=tmp=l=r=0;
        while(1){
            if(r>=n)    break;
            while(1){
                if(r>=n)    break;
                if(a[r]>=m){
                    tmp++;
                }
                if(tmp==k){
                    sum+=(n-r);
                    break;
                }
                r++;
            }
            if(r>=n)    break;
            while(1){
                if(l>r)    break;
                if(a[l]<m){
                    sum+=(n-r);
                }
                else{
                    l++;
                    tmp--;
                    break;
                }
                l++;
            }
            r++;
        }
        printf("%lld\n",sum);
    }
    return 0;
}

Other people's code ：

Code 1：

#include <cstdio>
#define N 200050
int a[N];
long long d[N];
int main()
{
    int T,n,m,k;
    scanf("%d",&T);
    while(T--){
        scanf("%d %d %d",&n,&m,&k);
        int t=0;
        long long sum=0;
        for(int i=1; i<=n; i++){
            scanf("%d",&a[i]);
            if(a[i]>=m){
                d[++t]=i;
                if(t>=k){
                    if(t==k) sum+=d[1]*(n-i+1);
                    else sum+=(d[t-k+1]-d[t-k])*(n-i+1);
                }
            }
        }
        printf("%lld\n",sum);
    }
    return 0;
}

Code 2：

#include <cstdio>
#define N 200050
int a[N];
int main(){
    int T,n,m,k;
    scanf("%d",&T);
    while(T--){
        scanf("%d %d %d",&n,&m,&k);
        int t=0;
        long long sum=0;
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        int r=1,num=0;
        for(int i=1;i<=n;i++){
            while(r<=n&&num<k){
                if(a[r]>=m) num++;
                r++;
            }
            if(num>=k) sum+=n-r+2;
            if(a[i]>=m) num--;
        }
        printf("%lld\n",sum);
    }
    return 0;
}

Code 3：

#include <cstdio>
const int MAXN = 200000 + 9;
int a[MAXN];
void solve(){
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    long long ans = 0;
    int r;
    int t = 0;
    for (r = 0; r < n; r++) {
        if (a[r] >= m) t++;
        if (t == k) break;
    }
    ans += (n - r);
    for (int l = 1; l < n; l++) {
        if (a[l - 1] >= m) {
            t--;
            r++;
            for (;r < n; r++) {
                if (a[r] >= m) t++;
                if (t == k) break;
            }
            if (r >= n) break;
            ans += (n - r);
        } else {
            ans += (n - r);
        }
    }
    cout << ans << endl;
}
int main(){
    //freopen("in", "r", stdin);
    int t;
    scanf("%d",&t);
    while(t--){
        solve();
    }
}