Can you find it?

Posted 2020-08-15 勿忘初心0924

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 1140 Accepted Submission(s): 370

Problem 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

Author wangye

Source HDU 2007-11 Programming Contest

Recommend 威士忌

/*
二分查找嘛
第一遍想的在三个数组整合到一起查找，但是超内存，想了一下，只整合两个数组的话，时间上虽然复杂了，但是内存小了
*/
#include<bits/stdc++.h>
#define MAX 505
using namespace std;
long long a[MAX],b[MAX],c[MAX],d[MAX*MAX];
int BinarySearch(long long num[],long long end,long long n)/*二分查找*/
{
    long long l=0,r=end,mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(num[mid]==n)
            return 1;
        if(num[mid]>n)
            r=mid-1;
        else if(num[mid]<n)
            l=mid+1;
    }
    if(num[l]==n)
        return 1;
    return 0;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    long long L,N,M,t,n;
    int Case=1;
    while(scanf("%lld%lld%lld",&L,&N,&M)!=EOF)
    {
        for(int i=0;i<L;i++)
            scanf("%lld",&a[i]);
        for(int i=0;i<N;i++)
            scanf("%lld",&b[i]);
        for(int i=0;i<M;i++)
            scanf("%lld",&c[i]);
        long long len=0;
        for(int i=0;i<L;i++)
            for(int j=0;j<N;j++)
                    d[len++]=a[i]+b[j];
        sort(d,d+len);
        scanf("%lld",&t);
        printf("Case %d:\n",Case++);
        while(t--)
        {
            scanf("%lld",&n);
            int f=0;
            for(int i=0;i<M;i++)
            {
                if(BinarySearch(d,len,n-c[i]))
                {
                    f=1;
                    break;
                }    
            }
            if(f)
                puts("YES");
            else
                puts("NO");
        }
    }
}