首页 > 代码库 > hdu 2141 Can you find it?

hdu 2141 Can you find it?

2024-07-02 07:47:41   228人阅读

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 9863    Accepted Submission(s): 2587


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


主要是把两个数组合并为一个数，这一点没想起来。

然后二分查找，因为只需要判断能否找到该元素，所以一般的二分查找就行。


#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#define N 505
int a[N],b[N],c[N];
int ab[N*N];
int cmp(const void*a,const void*b)
{
	return *(int*)a-*(int*)b;
}
int find(int x,int high)         //二分查找
{
	int mid,low=0;
	while(low<=high)
	{
		mid=(low+high)/2;
		if(ab[mid]>x)
			high=mid-1;
		else if(ab[mid]<x)
			low=mid+1;
		else
			return 1;
	}
	return 0;
}
int main()
{
	int l,m,n;
	int s,q,i,j,k,cnt=1;
	while(scanf("%d%d%d",&l,&m,&n)!=-1)
	{
		for(i=0;i<l;i++)
			scanf("%d",&a[i]);
		for(i=0;i<m;i++)
			scanf("%d",&b[i]);
		for(i=0;i<n;i++)
			scanf("%d",&c[i]);
		for(i=0,k=0;i<l;i++)      //把两个数组和为一个数组
		{
			for(j=0;j<m;j++)
				ab[k++]=a[i]+b[j];
		}
		qsort(ab,k,sizeof(ab[0]),cmp);
		qsort(c,n,sizeof(c[0]),cmp);
		scanf("%d",&q);
		printf("Case %d:\n",cnt++);
		while(q--)
		{
			scanf("%d",&s);
			int flag=0;
			for(i=0;i<n;i++)
			{
				if(find(s-c[i],k))
				{
					flag=1;
					break;
				}
			}
			if(flag)
				printf("YES\n");
			else
				printf("NO\n");
		}
	}
	return 0;
}


联系
我们