hdoj_2141

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 6083    Accepted Submission(s): 1585

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
二分水题，开始没有预先保存a[i]+b[j]，wa。也算是个小小的优化吧
#include <iostream>#include <cstdio>#include <algorithm>using namespace std;int main(){freopen("in.txt","r",stdin);int a[505],b[505],c[505],sum[250050];int l,n,m,s,i,x,j;int ans = 0;int cnt = 1;int k = 0;while(scanf("%d %d %d",&l,&n,&m)!=EOF){memset(a,0,sizeof(a));memset(b,0,sizeof(b));memset(c,0,sizeof(c));memset(sum,0,sizeof(sum));for(i=0;i<l;i++)scanf("%d",&a[i]);for(i=0;i<n;i++)scanf("%d",&b[i]);for(i=0;i<m;i++)scanf("%d",&c[i]);k = 0;for(i=0;i<l;i++){for(j=0;j<n;j++){sum[k++] = a[i] + b[j];}}sort(sum,sum+k);sort(c,c+m);scanf("%d",&s);printf("Case %d:\n",cnt++);while(s--){ans = 0;scanf("%d",&x);for(i=0;i<m;i++){int left = 0;int right = l * n - 1;while(left<=right){int mid = (left + right) / 2;if(sum[mid] + c[i] > x){right = mid - 1;}else if(sum[mid] + c[i] < x){left = mid + 1;}else {printf("YES\n");ans = 1;break;}}if(ans==1) break;}if(ans==0) printf("NO\n");}}return 0;}