Can you find it?+合并字串+二分查找-优快云博客

本文介绍了一个经典的算法问题——给定三个整数序列A、B、C及目标值X，判断是否存在A中的元素Ai、B中的元素Bj与C中的元素Ck，使得它们的和等于X。通过合并两个序列并使用二分查找的方法来提高求解效率。

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Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 7114 Accepted Submission(s): 1851

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

Sample Output

  
   Case 1:
NO
YES
NO

Author

wangye

Source

HDU 2007-11 Programming Contest

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/*
思路：二分查找
技巧：合并其中两个串，再进行二分，不然超时
*/

#include<iostream>
using namespace std;
int d[250010];
int cmp(const void *a,const void *b){
	return *(int *)a-*(int *)b;
}
int main()
{
	int l,n,m,count=1;
	while(cin>>l>>n>>m){
		int a[502],b[502],c[502];
		int i,j;
		for(i=0;i<l;i++)
			cin>>a[i];
		for(i=0;i<n;i++)
			cin>>b[i];
		for(i=0;i<m;i++)
			cin>>c[i];
		int k=0;
		for(i=0;i<n;i++)
			for(j=0;j<m;j++)
				d[k++]=b[i]+c[j];
			
			qsort(d,k,sizeof(d[0]),cmp);
			int s;
			cin>>s;
			cout<<"Case "<<count++<<":"<<endl;
			while(s--){
				int x,flag=0;
				cin>>x;
				for(i=0;i<l;i++){
					int num=x-(a[i]);
					int low=0,high=k,mid;
					while(low<=high){
						mid=(low+high)/2;
						if(d[mid]<num)
							low=mid+1;
						else
						{
							if(d[mid]==num)
								break;
							else
								high=mid-1;
						}
					}
					if(num==d[mid]){
						flag=1;
						cout<<"YES"<<endl;
						break;
					}
				}
				if(!flag)
					cout<<"NO"<<endl;
			}
			
	}
	return 0;
}