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三数求和算法实现

最新推荐文章于 2022-01-28 18:03:20 发布

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#二分搜索

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本文介绍了一种解决三数求和问题的有效算法。通过预处理两个数的和并进行排序，结合第三个数使用二分查找的方式快速判断是否存在三个数之和等于目标值X。此方法适用于大量查询的情况。

Description
Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

Input
There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

Output
For each case, firstly you have to print the case number as the form “Case d:”, then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print “YES”, otherwise print “NO”.

Sample Input

3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10

Sample Output

Case 1:
NO
YES
NO

求三个数表中是否有满足和为X的三个数，先求前两个数表的和并排序，再枚举第三个数表的数字，用二分搜索法查找和数组里面是否有满足条件的数字，不需要查重。

#include <iostream>
#include <algorithm>
using namespace std;
#define MAXN 505

int a[MAXN];
int b[MAXN];
int c[MAXN];
int d[MAXN * MAXN];

int bs(int key, int num)
{
    int lo=0,hi=num;
    int mi;
    while(lo<=hi)
    {
        mi=((hi-lo)>>1)+lo;//lo和hi比较大的时候相加可能爆int
        if(d[mi]==key) return mi;
        else if(d[mi]<key) lo=mi+1;
        else hi=mi-1;
    }
    return -1;//未找到
}

int main()
{
    int t = 1;
    int l,n,m;
    int s,key;
    int i,j;
    int isYes = 0;
    while(cin >> l >> n >> m)
    {
        for(i = 0; i < l; i++)
            cin >> a[i];
        for(i = 0; i < n; i++)
            cin >> b[i];
        for(i = 0; i < m; i++)
            cin >> c[i];
        cin >> s;
        for(int i = 0; i < l; i++)
            for(int j = 0; j < n; j++)
                d[i*n+j] = a[i] + b[j];
        sort(d, d+l*n);
        cout << "Case " << t << ":" << endl;
        t++;
        while(s--)
        {
            cin >> key;
            for(i = 0; i < m; i++)
            {
                if(bs(key - c[i], l * n) != -1)
                {
                    isYes = 1;
                    cout << "YES" << endl;
                    break;
                }
            }
            if(isYes != 1)
                cout << "NO" << endl;
            isYes = 0;
        }
    }
    return 0;
}