Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 9266 Accepted Submission(s): 2418
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 3
1 2 3
1 2 3
1 2 3
3
1 4 10
Sample Output
Case 1:
NO
YES
NO
Author
wangye
Source
Recommend
1 /* 2 题意：Ai+Bj+Ck = X; 3 其中Ai 属于数组a[ ] , Bj 属于数组b[ ], Ck属于数组 C[ ] 的元素。 4 求是否能找到满足X。而且X个数有1000，A,B,C个数为500. 5 思路1. 枚举 X-Ck ，Bj，用二分找Ai是否存在。 6 时间复杂度 f( 1000*500*500*log(500) ) 7 思路2. Ai+Bj,看成一项。用数组保存tom[ 500*500+1]; 8 枚举 X-Ck，用二分找tom[ ]; 9 时间复杂度 f( 1000*500*log(500*500) ) 10 11 */ 12 13 #include<iostream> 14 #include<stdio.h> 15 #include<cstring> 16 #include<cstdlib> 17 #include<algorithm> 18 using namespace std; 19 20 int l[502],n[502],m[502]; 21 int a[250003]; 22 23 int EF(int l,int r,int x) 24 { 25 int mid; 26 while(l<=r) // is not while(l<r) !! 27 { 28 mid=(l+r)/2; 29 if( a[mid]>x) 30 r=mid-1; 31 else if( a[mid]<x) 32 l=mid+1; 33 else return 1; 34 } 35 return -1; 36 } 37 int main() 38 { 39 int i,j; 40 int L,N,M,S,num=0,x,k,len,t; 41 while(scanf("%d%d%d",&L,&N,&M)>0) 42 { 43 for(i=1;i<=L;i++) 44 scanf("%d",&l[i]); 45 for(i=1;i<=N;i++) 46 scanf("%d",&n[i]); 47 for(i=1;i<=M;i++) 48 scanf("%d",&m[i]); 49 len=L*N;t=0; 50 for(i=1;i<=L;i++) 51 for(j=1;j<=N;j++) 52 { 53 a[++t]=l[i]+n[j]; 54 } 55 sort(a+1,a+1+len); 56 scanf("%d",&S); 57 printf("Case %d: ",++num); 58 while(S--) 59 { 60 scanf("%d",&x); 61 for(i=1;i<=M;i++) 62 { 63 k=EF(1,len,x-m[i]); 64 if( k==1 ) break; 65 } 66 if( k==-1) printf("NO "); 67 else printf("YES "); 68 } 69 } 70 return 0; 71 }