HDUoj 2141 Can you find it?(排序+二分)

 Can you find it?
Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 26449    Accepted Submission(s): 6673

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
#include<stdio.h>#include<algorithm>using namespace std;#define X 500int M,L,N,S;int a[X+5],b[X+5],c[X+5],sum[X*X+5];bool find(int x){int low = 0, high = N*L;while(high-1>=low){int mid = (low+high)/2;if(sum[mid]==x)return true;else if((sum[mid]<x))low = mid + 1;elsehigh = mid;}return false;}int main(){int x,k=0;while(~scanf("%d%d%d",&L,&N,&M)){k++;for(int i = 0;i < L;i ++)scanf("%d",&a[i]);for(int i = 0;i < N;i ++)scanf("%d",&b[i]);for(int i = 0;i < M;i ++)scanf("%d",&c[i]);for(int i = 0;i < L;i ++){for(int j = 0;j < N;j ++){sum[i*N+j] = a[i] + b[j];}}scanf("%d",&S);sort(sum,sum+N*L);printf("Case %d:\n",k);for(int i = 0;i < S;i ++){int flag = 0;scanf("%d",&x);for(int j = 0;j < M;j ++){if(find(x-c[j])){flag = 1;break;}}if(flag)printf("YES\n");elseprintf("NO\n");}}return 0;}