C语言排序(12)___Can you find it?(Hdu 2141)

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
   
   
   
   

 

题目大意：

输入数据第一排是三个数，分别表示三个数组的大小. (三个数组的大小的最大值都是500)

接下来的三排是输入三个数组的数字.

然后第四排输入一个整数n表示接下来n排测试数据.

每个测试数据是一个整数m.

要求在上面的三个数组中每个数组找一个数(共三个数)相加之和等于该测试数据m.如果找到了就输出YES,如果不存在这样的三个数那么就输出NO.

大致思路：

因为题目给的数据每个数组最多有500个元素,那么如果三个for循环就是500*500*500肯定超时,(姑且我们这里把第一第二第三数组叫为a,b,c数组)

这个时候我们可以把c数组排序,然后用两个for循环枚举a,b两个数组中所有组合的和k,然后再用二分查找在c数组中查找m-k这个数.这样时间最多为500*500*log500≈ 2250000

或者是在a,b两个数组中所有组合的和k组成的数组中去二分查找m-c[i]这个数(以下代码便是).

代码如下：

#include <stdio.h>
#include <algorithm>
using namespace std;
int a[501],b[501],m[250001],c[501],flag;
void find(int a[],int b,int n)
{
    int left,right,mid;
    left=0,right=n-1;
    while(left<=right)
    {
        mid=(left+right)/2;
        if(a[mid]==b)
        {
            flag=1;
            break;
        }
        else if(a[mid]>b)
            right=mid-1;
        else
            left=mid+1;
    }
}
int main()
{
    int i,j,k,n1,n2,n3,n4,ci=0;
    while(scanf("%d%d%d",&n1,&n2,&n3)!=EOF)
    {
        ci++;
        for(i=0;i<n1;i++)
            scanf("%d",&a[i]);
        for(i=0;i<n2;i++)
            scanf("%d",&b[i]);
        for(i=0;i<n3;i++)
            scanf("%d",&c[i]);
        n4=0;
        for(i=0;i<n1;i++)
            for(j=0;j<n2;j++)
                m[n4++]=a[i]+b[j];
        sort(m,m+n4);
        int N;
        scanf("%d",&N);
        printf("Case %d:\n",ci);
        while(N--)
        {
            flag=0;
            scanf("%d",&k);
            for(i=0;i<n3;i++)
            {
                find(m,k-c[i],n4);
                if(flag==1)
                    break;
            }
            if(flag==1)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    return 0;
}