Costume Party

Time Limit: 1000MS Memory Limit: 65536K

Description

It’s Halloween! Farmer John is taking the cows to a costume party, but unfortunately he only has one costume. The costume fits precisely two cows with a length of S (1 ≤ S ≤ 1,000,000). FJ has N cows (2 ≤ N ≤ 20,000) conveniently numbered 1…N; cow i has length Li (1 ≤ Li ≤ 1,000,000). Two cows can fit into the costume if the sum of their lengths is no greater than the length of the costume. FJ wants to know how many pairs of two distinct cows will fit into the costume.

Input

  • Line 1: Two space-separated integers: N and S
  • Lines 2…N+1: Line i+1 contains a single integer: Li

Output

  • Line 1: A single integer representing the number of pairs of cows FJ can choose. Note that the order of the two cows does not matter.

Sample Input

4 6
3
5
2
1

Sample Output

4

思路：

本题可以直接暴力，也可以用二分查找的方法找到不超过m的下标，然后就可以计算出来最后的总和了。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 20010;
int main() {
    ios::sync_with_stdio(false);
    int n, sum = 0, m;
    int a[maxn];
    scanf("%d %d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    sort(a, a + n);
    for (int i = 0; i < n; i++) {
        int x = upper_bound(a + i + 1, a + n, m - a[i]) - a - i - 1;
        if (x == 0) break;
        sum += x;
    }
    printf("%d", sum);
    return 0;
}