Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 19416 Accepted Submission(s): 4891
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
1 2 3
1 2 3
1 2 3
3
1
4
10
Sample Output
Case 1:
NO
YES
NO
lower_bound提交就会出错，binary_search()也可以过
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <algorithm> 5 #include <set> 6 using namespace std; 7 int a[502],b[502],c[502],x; 8 __int64 su[502*502]; 9 int l,n,m,s,u; 10 int d=1,sum; 11 bool solve(int x) 12 { 13 int i,k; 14 int ans; 15 for(i=0;i<m;i++) 16 { 17 ans=x-c[i]; 18 int l=0,r=u-1,mid,e; 19 while(l<=r) 20 { 21 mid=(l+r)/2; 22 if(su[mid]==ans) return 1; 23 else if(su[mid]>ans) r=mid-1; 24 else l=mid+1; 25 } 26 } 27 return 0; 28 } 29 int main() 30 { 31 int i,j; 32 freopen("in.txt","r",stdin); 33 while(scanf("%d%d%d",&l,&n,&m)!=EOF) 34 { 35 for(i=0;i<l;i++) scanf("%d",&a[i]); 36 for(i=0;i<n;i++) scanf("%d",&b[i]); 37 for(i=0;i<m;i++) scanf("%d",&c[i]); 38 scanf("%d",&s); 39 u=0; 40 for(i=0;i<l;i++) 41 for(j=0;j<n;j++) 42 su[u++]=a[i]+b[j]; 43 sort(su,su+u); 44 printf("Case %d: ",d++); 45 for(i=0;i<s;i++) 46 { 47 scanf("%d",&x); 48 if(solve(x)) printf("YES "); 49 else printf("NO "); 50 } 51 } 52 }