hdu-2141Can you find it?（二分搜索求和）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 24425    Accepted Submission(s): 6186

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

 

Recommend

威士忌

 

先把第一组和第二组所以加起来的可能性都保存起来 在从小到大排序 k=sum-c[i]  二分搜索能否找到k 找到即可以输出yes

#include<iostream>#include<string>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;int a[600],b[600],c[600],sum[360000];int n,m,l,s;int se(int num,int k){int l=k-1,r=0;int mid;while(l>=r){mid=(l+r)/2;if(sum[mid]==num)return 1;else if(sum[mid]>num)l=mid-1;else r=mid+1;}return 0;}int main(){int cnt=1;while(~scanf("%d%d%d",&l,&n,&m)){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]);int k=0;for(int i=0;i<l;i++){for(int j=0;j<n;j++){sum[k++]=a[i]+b[j];}}sort(sum,sum+k);scanf("%d", &s);  printf("Case %d:\n", cnt++);  int x;  while(s--)  {  scanf("%d",&x);  int f=0;  for(int i=0;i<m&&f==0;i++)  {  int num=(x-c[i]);  if(se(num,k))  {  f=1;  }  }  if(f)printf("YES\n");  else printf("NO\n");  }}return 0;}