P1102 A-B数对题解

题目传送门

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int N = 200010;
int a[N];
int n, c;
LL cnt; //累加和不加long long 死翘翘

int main() {
    //输入
    cin >> n >> c;
    for (int i = 1; i <= n; i++) cin >> a[i];

    //排序
    sort(a + 1, a + 1 + n);

    //遍历每一个数字
    for (int i = 1; i < n; i++) {
        //每个数字可能都是多个，我们不能漏掉每个数字
        int left = 0, right = 0;

        //左端点
        int l = 1, r = n, k = a[i] + c;//k为目标值
        while (l < r) {
            int mid = (l + r) >> 1;
            if (a[mid] >= k)r = mid;
            else l = mid + 1;
        }
        if (a[l] == k)left = l;

        //右端点
        l = 1, r = n;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (a[mid] <= k)l = mid;
            else r = mid - 1;
        }
        if (a[l] == k) right = r;

        //左右端点差就是个数,其实left如果不是0，那么right肯定不是0
        if (left && right) cnt += right - left + 1;
    }
    //输出结果
    printf("%lld", cnt);
    return 0;
}

标签：P1102,int,题解,LL,数对,mid,long,while,left
来源： https://www.cnblogs.com/littlehb/p/15042728.html