HDU 2141 Can you find it? （二分）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 22628    Accepted Submission(s): 5724

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

Author

wangye

Source

HDU 2007-11 Programming Contest

 

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题解：给你三组A、B、C的值和S值，问是否能找到Ai、Bj、Ck，使得Ai+Bj+Ck = S。
           二分，先将a数组和b数组，再相加成一个数组ab[500*500]。这样就相当于ab[i] + c[j] = s。
           再变形一下， ab[i] = s - c[j].
         只要在ab数组中用二分查找是否存在s-c[j]就可以了。

AC代码：

#include<iostream>#include<algorithm>#include<stdio.h>using namespace std;int a[505],b[505],c[505];int ab[250005];bool Binary_Search( int a, int l , int R ){//二分   if( l > R )     return false;   int mid = ( l + R ) / 2;    if (ab[mid] == a )    return true;   else if( ab[mid] > a)    Binary_Search( a, l, mid-1);   else    Binary_Search( a, mid+1, R);}int main(){ int l,n,m; int i,j,k; int s,sum,cnt = 1; while( cin >> l >> n >> m ) {  for( i = 0; i < l; i++)   cin >> a[i];     for( i = 0; i < n; i++)   cin >> b[i];     for( i = 0; i < m; i++)    cin >> c[i];     for(k=0, i=0;i<l; i++)   {   for( j=0; j<n; j++)            {            ab[k++] = a[i] + b[j];}   }      sort( ab, ab + k );  cin >> s;  cout << "Case " << cnt++ << ":"<<"\n";  for( i = 0; i < s; i++ )  {   cin >> sum;   for( j = 0; j < m; j++)    if ( Binary_Search(sum-c[j] , 0 , k-1) ) //查找是否有满足的和     break;   if( j == m)    puts("NO");    else    puts("YES");  }} return 0;}