51nod 1001数组中等于k的数对

题目

给n个不重复的数，问有多少数对和等于k

分析

两种方法

1. 对于每一个都二分找
2. 两头逼近找数对

#include <bits/stdc++.h>
#pragma GCC diagnostic error "-std=c++11"
#define d(x) cout << (x) << endl
#define INF 0x3f3f3f3f
#define eps 1e-8
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 5e4 + 10;

int k, n;
int a[N];
int find(int l, int r, int num)
{
    while(l <= r){
        int mid = (l + r) >> 1;
        if(num == a[mid])
            return 1;
        else if(num > a[mid]){
            l = mid+1;
        }else{
            r = mid-1;
        }
    }
    return 0;
}
int main()
{
    scanf("%d%d", &k, &n);
    for (int i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    sort(a, a + n);
    int ans = 0;
    int tag = 1;
    for (int i = 0; i < n - 1; i++){
        if(find(i+1, n-1, k-a[i])){
            tag = 0;
            printf("%d %d\n", a[i], k - a[i]);
        }
    }
    if(tag)
        printf("No Solution\n");
    return 0;
}

#include <bits/stdc++.h>
#pragma GCC diagnostic error "-std=c++11"
#define d(x) cout << (x) << endl
#define INF 0x3f3f3f3f
#define eps 1e-8
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 5e4 + 10;

int k, n;
int a[N];
int main()
{
    scanf("%d%d", &k, &n);
    for (int i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    sort(a, a + n);
    int tag = 1;
    int l = 0;
    int r = n - 1;
    while(l < r){
        int mid = a[l] + a[r];
        if(mid == k){
            printf("%d %d\n", a[l++], a[r--]);
            tag = 0;
        }else if(mid < k){
            l++;
        }else{
            r--;
        }
    }
    if(tag)
        puts("No Solution");
    return 0;
}