Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 42520    Accepted Submission(s): 10315

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

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今天从基础开始学起，以为二分挺容易的，但是忽视了题目需要考虑时间复杂度，会不会爆int，等等。

存不存在ai,bi,ci加起来是x。

如果直接枚举那也太简单了吧，我就想着，大循环a，小循环b，小小循环对c二分，发现这样也超时！！！

然后，我就大循环a，小循环对b二分，小小循环对c二分，结果这样的思路是完全错的！！！

然后看看题解，要先把a+b的所有可能都存起来放到sum里，再大循环c，对sum二分。开始还觉得这个思路不是和我一开始的思路差不多吗？仔细一想，我大循环a，小循环b，一方面会有很多重复的a+b，重复带入算，费时，另一方面，每输入一个x，又要大循环啊，小循环b的，很费时。

#include <iostream>
#include <stack>
#include <string.h>
#include <stdio.h>
#include<queue>
#include<algorithm>
#define ll long long
using namespace std;
int a[505];
int b[505];
int c[505];
int sum[500005];
int main()
{
    int L,N,M;
    int k=0;
    while(~scanf("%d %d %d",&L,&N,&M))
    {
        k++;
        for(int i=1;i<=L;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=N;i++)
        {
            scanf("%d",&b[i]);
        }
        for(int i=1;i<=M;i++)
        {
            scanf("%d",&c[i]);
        }
        int n,x;
        printf("Case %d:\n",k);
        scanf("%d",&n);
        int p=1;

        for(int i=1;i<=L;i++)
        {
            for(int j=1;j<=N;j++)
            {
                sum[p++]=a[i]+b[j];
            }
        }
        sort(sum+1,sum+p);
        sort(c+1,c+1+M);
        while(n--)
        {
            scanf("%d",&x);
            bool f=0;
            for(int i=1;i<=M;i++)
            {
                if(sum[1]>x-c[i])//不能写成sum[1]+c[1]>x，因为可能会爆int
                    break;
                int le=1;int ri=p-1;
                while(le<=ri)
                {
                    int mid=(le+ri)/2;
                    if(sum[mid]==x-c[i])
                    {
                        f=1;
                        break;
                    }
                    else if(sum[mid]>x-c[i])
                    {
                        ri=mid-1;
                    }
                    else if(sum[mid]<x-c[i])
                        le=mid+1;
                }
                if(f)
                    break;
            }
            if(f)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    return 0;
}       

原文地址：https://www.cnblogs.com/caiyishuai/p/9527220.html