hdu杭电 2141 Can you find it? 【二分 N*logN】

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 31 2 31 2 31 2 331410

 

Sample Output

Case 1:NOYESNO

 

//ac代码

//Can you find it#include<cstdio>#include<algorithm>using namespace std;int s[550*550];int main(){int m,n,l;//表示abc int t,x;int i,k,j;int mid;int a[550],b[550],c[550];int num=0;while(~scanf("%d%d%d",&m,&n,&l)){for(i=0;i<m;++i){scanf("%d",a+i);}for(i=0;i<n;++i){scanf("%d",b+i);}for(i=0;i<l;++i){scanf("%d",c+i);}int p=0;for(j=0;j<n;++j)//把b集合和c集合的数存到s数组中{for(k=0;k<l;++k){s[p++]=b[j]+c[k];}}sort(s,s+p);//排序printf("Case %d:\n",++num);scanf("%d",&t);while(t--){int flag=0;scanf("%d",&x);for(i=0;i<m;++i){int l,r;if(s[0]+a[i]==x||s[p-1]+a[i]==x){flag=1;break;}/*else if(s[0]+a[i]>x || s[p-1]+a[i]<x){continue；}*/l=0;r=p-1;while(r>=l){mid=(l+r)/2;if(s[mid]+a[i]==x){flag=1;break;}else if(s[mid]+a[i]<x)  l=mid+1;else if(s[mid]+a[i]>x)  r=mid-1;}if(flag) break;}if(flag) printf("YES\n");else printf("NO\n");}} return 0;}