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最新推荐文章于 2023-04-16 15:29:27 发布

原创最新推荐文章于 2023-04-16 15:29:27 发布 · 251 阅读

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本文介绍了一种解决三数组求和问题的有效算法。该问题要求从三个数组中各选取一个元素，判断它们的和是否等于给定的目标值。文章通过双重循环结合二分查找的方法，实现了高效求解。

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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.

InputThere 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.
OutputFor 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

Sample Output

Case 1:
NO
YES
NO

题意：分别有l,n,m长度的串，从每个串中取一个数相加，若能得到给出数的和，则YES，否则NO

思路：刚开始写的三层for循环，然后map存三个串能得到的所有和，然后爆内存……

正确的思路是先将前两个for循环求出和得到一个新的数组，sort排序，遍历第三个串，从得到的新串中二分寻找答案

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
using namespace std;
const int MAXN=500+10;
//const int inf=0x3f3f3f;
long long x[MAXN],y[MAXN],z[MAXN];
long long s[MAXN*MAXN];
int l,n,m;
int judge(long long p,long long answer,int left,int right)
{
    while(left<=right)
    {
        int mid=(left+right)/2;
        if(s[mid]+p==answer)
            return 1;
        else if(s[mid]+p<answer)
            left=mid+1;
        else if(s[mid]+p>answer)
            right=mid-1;
    }
    return 0;
}
int main()
{
    int t=0;
    while(~scanf("%d%d%d",&l,&n,&m))
    {
        for(int i=0; i<l; i++)
            scanf("%lld",&x[i]);

        for(int i=0; i<n; i++)
            scanf("%lld",&y[i]);

        for(int i=0; i<m; i++)
            scanf("%lld",&z[i]);
        int k=0;
        memset(s,0,sizeof(s));
        long long ans;
        for(int i=0; i<l; i++)
        {
            for(int j=0; j<n; j++)
            {
                s[k++]=x[i]+y[j];
            }
        }
        sort(s,s+k);
        int p;
        scanf("%d",&p);
        long long a;
        printf("Case %d:\n",++t);
        while(p--)
        {
            scanf("%lld",&a);
            int flag=0;
            for(int i=0; i<m; i++)
            {
                if(judge(z[i],a,0,k-1))
                {
                    flag=1;
                    break;
                }
            }
            if(!flag) printf("NO\n");
            else if(flag==1) printf("YES\n");
        }
    }
}