HDUoj 2141 Can you find it?(排序+二分)

本文链接：https://blog.youkuaiyun.com/wang2332/article/details/53905001

本文介绍了一道经典的算法题目——三数求和问题。该问题要求从三个已知序列中找出是否存在三项之和等于指定数值X。通过构建算法模型，采用排序与二分查找相结合的方法高效解决此问题。

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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): 26449 Accepted Submission(s): 6673

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

#include<stdio.h>
#include<algorithm>
using namespace std;
#define X 500
int M,L,N,S;

int a[X+5],b[X+5],c[X+5],sum[X*X+5];

bool find(int x)
{
	int low = 0, high = N*L;
	while(high-1>=low){
		int mid = (low+high)/2;
		if(sum[mid]==x)
			return true;
		else if((sum[mid]<x))	
			low = mid + 1;
		else
			high = mid;
	}
return false;
}
int main()
{
	int x,k=0;
	while(~scanf("%d%d%d",&L,&N,&M)){
		k++;
		for(int i = 0;i < L;i ++)
			scanf("%d",&a[i]);
		for(int i = 0;i < N;i ++)
			scanf("%d",&b[i]);
		for(int i = 0;i < M;i ++)
			scanf("%d",&c[i]);
		for(int i = 0;i < L;i ++){
			for(int j = 0;j < N;j ++){
				sum[i*N+j] = a[i] + b[j];
			}
		}
		scanf("%d",&S);
		sort(sum,sum+N*L);
		printf("Case %d:\n",k);
		for(int i = 0;i < S;i ++){
			int flag = 0;
			scanf("%d",&x);
			for(int j = 0;j < M;j ++){
				if(find(x-c[j])){
					flag = 1;
					break;
				}
			}
			if(flag)
				printf("YES\n");
			else
				printf("NO\n");
		}
		
	}
return 0;
}