1.二分查找B = A - C

#include<bits/stdc++.h>
using namespace std;

const int N = 200010;
long long q[N], c, res;
int n;

int main()
{
    cin >> n >> c;
    for(int i = 0; i < n; i ++ ) scanf("%d", &q[i]);

    sort(q, q + n);


    for(int i = 0; i < n; i ++ ){

        if(q[i] <= c) continue;

        int l = 0, r = i - 1, x = q[i] - c;

        while(l < r){
            int mid = (l + r) / 2;
            if(q[mid] >= x) r = mid;
            else l = mid + 1;
        }

        if(q[l] != x) continue;

        int lr = l;
        r = i - 1;
        while(l < r){
            int mid = (l + r + 1) / 2;
            if(q[mid] <= x) l = mid;
            else r = mid - 1;
        }

        // printf("%d %d %d\n", x, lr, l);
        res = res + r - lr + 1;

    }
    cout << res;
}

2.哈希(手写开放寻址法有一个测试点tle)

#include<bits/stdc++.h>
using namespace std;

const int N = 200010;
long long c, q[N], res;
int n;
map<long long, long long> mp;


int main()
{
    cin >> n >> c;
    for(int i = 0; i < n; i ++ ){
        cin >> q[i];
        mp[q[i]] ++;
    }

    for(int i = 0; i < n; i ++ ){
        res += mp[q[i] - c];
    }
    cout << res;
}

3.离散化

#include<bits/stdc++.h>
using namespace std;

const int N = 200010;
long long c, q[N], s[N], res;
int n;
vector<long long> all;

int find(int x)
{
    int l = 0, r = all.size() - 1;
    while(l < r){
        int mid = (l + r) / 2;
        if(all[mid] >= x) r = mid;
        else l = mid + 1;
    }

    if(all[l] == x) return l;
    else return -1;
}

int main()
{
    cin >> n >> c;
    for(int i = 0; i < n; i ++ ){
        scanf("%lld", &q[i]);
        all.push_back(q[i]);
    }

    sort(all.begin(), all.end());
    all.erase(unique(all.begin(), all.end()), all.end());

    for(int i = 0; i < n; i ++ ) s[find(q[i])] ++;

    for(int i = 0; i < n; i ++){
        int t = s[find(q[i] - c)];
        if(t != -1) res += t;
    }
    cout << res;

}