用c语言或c++解决这个问题描述:描述:Given an array, find the number of [i,j] (i and j are the indexes of the array (数组下标), not values) tuples in the array that satisfy array[i]+array[j]<=k. 输入:There are several test cases. For each test case, The first line contains two positive integers n (1<=n<=10^5) and k (1<=k<=10^18). The second line contains n integers, and the ith integer represents the value of array[i](1<= array[i]<=10^9). 输出:For each test case, output an integer representing the answer.
时间: 2024-02-18 07:03:22 浏览: 25
下面是C++代码实现:
```cpp
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5 + 5;
int n;
long long k, a[N];
int main() {
while (cin >> n >> k) {
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a, a + n);
int i = 0, j = n - 1;
long long ans = 0;
while (i < j) {
if (a[i] + a[j] <= k) {
ans += j - i;
i++;
} else {
j--;
}
}
cout << ans << endl;
}
return 0;
}
```
算法思路:
首先,我们可以对数组进行排序,然后使用双指针法,从左右两端开始向中间扫描。假设当前左指针指向i,右指针指向j,那么如果a[i] + a[j] <= k,则对于任意的i' (i' > i),a[i'] + a[j] <= k,因为数组已经排序,所以a[i'] >= a[i],所以对于任意的i' (i' > i),a[i'] + a[j] <= k,满足要求的(i', j)的数量就是j - i。因此,我们可以累加满足条件的(i', j)的数量,然后将左指针i向右移动一位。否则,我们将右指针j向左移动一位。当i >= j时,算法结束。
时间复杂度为O(nlogn)。