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

Sample Output

Case 1:

YES

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[];

int b[];

int c[];

int sum[];

int main()

{

    int L,N,M;

    int k=;

    while(~scanf("%d %d %d",&L,&N,&M))

    {

        k++;

        for(int i=;i<=L;i++)

        {

            scanf("%d",&a[i]);

        }

        for(int i=;i<=N;i++)

        {

            scanf("%d",&b[i]);

        }

        for(int i=;i<=M;i++)

        {

            scanf("%d",&c[i]);

        }

        int n,x;

        printf("Case %d:\n",k);

        scanf("%d",&n);

        int p=;

        for(int i=;i<=L;i++)

        {

            for(int j=;j<=N;j++)

            {

                sum[p++]=a[i]+b[j];

            }

        }

        sort(sum+,sum+p);

        sort(c+,c++M);

        while(n--)

        {

            scanf("%d",&x);

            bool f=;

            for(int i=;i<=M;i++)

            {

                if(sum[]>x-c[i])//不能写成sum[1]+c[1]>x，因为可能会爆int

                    break;

                int le=;int ri=p-;

                while(le<=ri)

                {

                    int mid=(le+ri)/;

                    if(sum[mid]==x-c[i])

                    {

                        f=;

                        break;

                    }

                    else if(sum[mid]>x-c[i])

                    {

                        ri=mid-;

                    }

                    else if(sum[mid]<x-c[i])

                        le=mid+;

                }

                if(f)

                    break;

            }

            if(f)

                printf("YES\n");

            else

                printf("NO\n");

        }

    }

    return ;

}

巴特西

HDU 2143 Can you find it?（基础二分）

Can you find it?

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