HDOJ 题目2141 Can you find it?（二分搜索）

此博客介绍了一个涉及三个序列的数值问题，目标是在给定的公式下找到符合条件的数值组合。通过输入序列长度、序列元素及目标值，算法进行优化计算，输出满足条件的查询结果。

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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): 11882 Accepted Submission(s): 3094

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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ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int cmp(const void *a,const void *b)
{
	return *(int *)a-*(int *)b;
}
int sum[255000],k;
int fun(int x)
{
	int l,r,mid;
	l=0;r=k-1;
	while(l<=r)
	{
		mid=(l+r)/2;
		if(sum[mid]>x)
			r=mid-1;
		else
			if(sum[mid]<x)
				l=mid+1;
			else
			{
				return 1;
				break;
			}
	}
	return 0;
}
int main()
{
	int l,n,m,kcase=0;
	while(scanf("%d%d%d",&l,&n,&m)!=EOF)
	{
		int a[5500],b[5500],c[5500],i,j,cot,w=1;
		k=0;
		for(i=0;i<l;i++)
			scanf("%d",&a[i]);
		for(i=0;i<n;i++)
		{
			scanf("%d",&b[i]);
		}
		for(i=0;i<m;i++)
			scanf("%d",&c[i]);
		for(i=0;i<l;i++)
		{
			for(j=0;j<n;j++)
				sum[k++]=a[i]+b[j];
		}
		qsort(sum,k,sizeof(sum[0]),cmp);
		scanf("%d",&cot);
		printf("Case %d:\n",++kcase);
		while(cot--)
		{
			int x;
			w=1;
			scanf("%d",&x);
			for(i=0;i<m;i++)
			{
				if(fun(x-c[i]))
				{
					w=0;
					break;
				}
			}
			if(w)
				printf("NO\n");
			else
				printf("YES\n");
		}
	}
}