Day6 - cattle-off 102C

Links: https://ac.nowcoder.com/acm/contest/102/C
Source: Cattle-off network

Title Description

 We define a value of an interval is the second largest number of it's elements, and of course an interval has at least two elements.

Given an array A with n elements and a number k, can you find the value of the k ^th largest interval?

Enter a description:

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

For each test case : 

The first line contains two integer numbers n,k(2 ≤ n ≤ 10 ⁵,1 ≤ k ≤ n(n−1)/2), the number of test cases. 

The second lines contains n integers A _i(1 ≤ A _i ≤ 10 ⁹), the elements of array A.

 

Output Description:

For each test case print the value of the k

^th

 largest interval.

Example 1

Entry

copy

2
3 3
1 2 3
5 1
1 2 2 3 3

Export

copy

1
3

Explanation

For the sample input, there are three intervals.
Interval [1 2 3] has value 2.
Interval [2 3] has value 2.
Interval [1 2] has value 1.
So the 3

^rd

largest interval is [1 2] whose value is 1. 

ideas: seeking k-th largest, the present problem and POJ3579 similar dichotomy then greedy test, foot emulated, at least two elements, to maintain a queue of two elements, for example: 
. 1 2  . 3 4, 2, 3 select, select 1 or 2, the front, rear choose 3 or 4, that is, four kinds of

typedef long long LL;

const int maxm = 1e5+10;

int buf[maxm], n, q[maxm];
LL k;

bool check(int x) {
    LL sum = 0, front = 0, rear = 0, last = -1;
    for(int i = 0; i < n; ++i) {
        if(buf[i] >= x) q[front++] = i;
        if(front - rear > 1) {
            sum += (q[rear] - last) * (n - i);
            last = q[rear++];
        }
    }
    return sum >= k;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%lld", &n, &k);
        for(int i = 0; i < n; ++i)
            scanf("%d", &buf[i]);
        int l = 1, r = 1e9, ans, mid;
        while(l <= r) {
            mid = (l + r) >> 1;
            if(check(mid)) {
                ans = mid;
                l = mid + 1;
            } else
                r = mid - 1;
        }
        printf("%d\n", ans);
    }
    return 0;
}

View Code