hdu2141 Can you find it? （二分查找）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 19334    Accepted Submission(s): 4860

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

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest

 

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解析：我们先预处理出a数列与b数列能够形成的所有数，存储在q数组当中，然后对q数组进行排序。
           对于每一个X，我们枚举c[i]，然后在q数组中二分查找是否存在X-c[i]，若存在，则“YES”，否则“NO”。
代码：

#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn=5e2;
int a[maxn+10],b[maxn+10],c[maxn+10];
int q[maxn*maxn+10];

bool find(int x)
{
int l=1,r=q[0],mid;
while(l<=r)
{
mid=(l+r)>>1;
if(q[mid]==x)return 1;
if(q[mid]<x)l=mid+1;
else r=mid-1;
}
return 0;
}

int main()
{
freopen("1.in","r",stdin);

int sum=0,i,j,k,s,x;
while(scanf("%d%d%d",&a[0],&b[0],&c[0])==3)
{
printf("Case %d:\n",++sum);
for(i=1;i<=a[0];i++)scanf("%d",&a[i]);
for(i=1;i<=b[0];i++)scanf("%d",&b[i]);
for(i=1;i<=c[0];i++)scanf("%d",&c[i]);
for(q[0]=0,i=1;i<=a[0];i++)
for(j=1;j<=b[0];j++)q[++q[0]]=a[i]+b[j];
sort(q+1,q+1+q[0]);
k=q[0],q[0]=1;
for(i=2;i<=k;i++)if(q[i]!=q[i-1])q[++q[0]]=q[i];
k=c[0],c[0]=1;
for(i=2;i<=k;i++)if(c[i]!=c[i-1])c[++c[0]]=c[i];

scanf("%d",&s);
for(i=1;i<=s;i++)
{
scanf("%d",&x);
for(j=1;j<=c[0];j++)if(find(x-c[j]))break;
printf("%s\n",(j<=c[0])?"YES":"NO");
}
}
return 0;
}