Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 8506 Accepted Submission(s): 2216
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 //375MS 1228K 1325 B C++ 2 /* 3 4 题意： 5 给出三行数和Y，问每是否存在行中一个数令 6 Ai+Bj+Ck==Y 7 成立 8 9 二分法： 10 笨笨的分析错时间复杂度了 11 先将两组数据合并，然后排序，然后进行二分 12 时间复杂度：O(n*n*lgn)... 13 14 */ 15 #include<stdio.h> 16 #include<string.h> 17 #include<stdlib.h> 18 #define N 505 19 int a[N],b[N],c[N],d[N*N]; 20 int l,n,m; 21 int cmp(const void*a,const void*b) 22 { 23 return *(int*)a-*(int*)b; 24 } 25 int main(void) 26 { 27 int t,s; 28 int cas=1; 29 while(scanf("%d%d%d",&l,&n,&m)!=EOF) 30 { 31 for(int i=0;i<l;i++) scanf("%d",&a[i]); 32 for(int i=0;i<n;i++) scanf("%d",&b[i]); 33 for(int i=0;i<m;i++) scanf("%d",&c[i]); 34 int cnt=0; 35 for(int i=0;i<n;i++) 36 for(int j=0;j<m;j++) 37 d[cnt++]=b[i]+c[j]; 38 qsort(a,l,sizeof(a[0]),cmp); 39 qsort(d,cnt,sizeof(d[0]),cmp); 40 scanf("%d",&t); 41 printf("Case %d: ",cas++); 42 while(t--){ 43 int flag=0; 44 scanf("%d",&s); 45 for(int i=0;i<l;i++){ 46 int tl=0,tr=cnt-1; 47 int tt=s-a[i]; 48 while(tl<tr){ 49 int mid=(tl+tr)/2; 50 if(d[tl]==tt || d[tr]==tt || d[mid]==tt){ //注意此处判断 51 flag=1;break; 52 }else if(d[mid]<tt) 53 tl=mid+1; 54 else tr=mid-1; 55 } 56 if(flag) break; 57 } 58 if(flag) puts("YES"); 59 else puts("NO"); 60 } 61 } 62 return 0; 63 }