HDU - 2141 Can you find it?

最新推荐文章于 2020-03-25 15:16:25 发布

acm_1361677193

最新推荐文章于 2020-03-25 15:16:25 发布

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本文链接：https://blog.youkuaiyun.com/acm_1361677193/article/details/38499085

本文介绍了一个经典的三数组求和问题及其解决方案。该问题要求从三个整数序列中找到是否存在三个数之和等于给定的目标值。通过预处理将两个数组的和存储为新的数组，并利用二分查找的方法进行快速匹配，从而实现高效解决。

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Can you find it?

Time Limit: 3000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

[Submit] [Go Back] [Status]

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

Source

HDU 2007-11 Programming Contest

题目大意：分别给定三组A、B、C的值和S值，问是否能找到Ai、Bj、Ck，使得Ai+Bj+Ck = S。

将a数组和b数组先相加成一个数组sab[500*500]。这样就相当于sab[i] + c[j] = s。再变形一下， sab[i] = s - c[j].
这样只要在sab数组中用二分查找是否存在s-c[j]就可以了。

简单的二分题，代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=500+5;
int sab[maxn*maxn],a[maxn],b[maxn],c[maxn];
int find(int n,int l,int r)
{
if(l>r) return 0;
int mid=(l+r)/2;
if(sab[mid]==n) return 1;
if(sab[mid]>n) find(n,l,mid-1);
// if(sab[mid]<n) find(n,mid+1,r);//写这句就神奇的wa了，不知为什么。它跟楼下不一样吗？？？
else find(n,mid+1,r);
}

int main()
{
int l,n,m,i,j,k,s,sum,kase=0;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
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++)
sab[k++]=a[i]+b[j];
sort(sab,sab+k);//注意排好序
scanf("%d",&s);
printf("Case %d:\n",++kase);//Case 前面没有空格
for(i=0;i<s;i++)
{
scanf("%d",&sum);
for(j=0;j<m;j++)
if(find(sum-c[j],0,k-1)) break;
if(j==m) printf("NO\n");
else printf("YES\n");
}
}
return 0;
}