（找数字）HDU - 2141

最新推荐文章于 2021-03-28 20:39:18 发布

UCAS王小二

最新推荐文章于 2021-03-28 20:39:18 发布

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分类专栏：二分&&三分

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本文针对HDU-2141算法题目进行了解析，介绍了一种通过预处理和二分查找来降低时间复杂度的有效解法。通过对输入数据的巧妙处理，大幅提升了算法效率。

HDU - 2141

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

Sample Output

Case 1:
NO
YES
NO

题意：A，B，C三个集合，再给出一个数X，判断是否 Ai+Bj+Ck = X；

思路：代码可能很好写，超时才是坑点，看代码；

//注意时间复杂度 
#include<stdio.h>
#include<algorithm>
using namespace std;
int d[3000000];//m*n的范围 
int main()
{
	int a[505],b[505],c[505];
	int m,n,k,i,j,t,p,h,w=1,r;
	while(scanf("%d %d %d",&m,&n,&k)!=EOF)
	{
	      for(i=0;i<m;i++)
		    scanf("%d",&a[i]);
		for(i=0;i<n;i++)
		    scanf("%d",&b[i]);
		for(i=0;i<k;i++)
		    scanf("%d",&c[i]);
		r=0;
		for(i=0;i<m;i++)
		    for(j=0;j<n;j++)
		        d[r++]=a[i]+b[j];//算出a[i]和b[j]的和，减少复杂度的关键一步 ，避免后续代码循环
		sort(d,d+r);//一次排序代替a,b,c三个数组的排序，避免后续代码循环  
		scanf("%d",&t);
		printf("Case %d:\n",w++);
		while(t--)
		{    
		    bool flag=true;
			scanf("%d",&h);
			for(i=0;i<k;i++)//由于上面做出的优化，这里只需要一次循环就可以了 
			{
			    int left=0,right=r-1,mid;
		    	    while(right>=left)//二分的关键应用 
		    	    {
		    	     	mid=(left+right)/2;
		    	        if(c[i]+d[mid]==h)//a,b,c三个数组同 h 进行比较 
		    	        {
		    	        	flag=false;
		    	        	break;
						}
						else if(c[i]+d[mid]>h)
						{
							right=mid-1;
						}
						else if(c[i]+d[mid]<h)
						{
							left=mid+1;
						}
					}
					if(!flag)
					break;
			}  
			if(!flag)
		          printf("YES\n");
		      else 
		          printf("NO\n");
		}
	}
	return 0;
}