链接:http://acm.hdu.edu.cn/showproblem.php?pid=2141
Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 11503 Accepted Submission(s): 3021
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
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这题写的不爽,不写题解了,看代码吧
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <iostream> 5 #include <algorithm> 6 7 using namespace std; 8 9 long long sum[250005]; 10 int str1[505],str2[505],str3[505]; 11 12 int Table(int l,int n,int m) 13 { 14 15 int cas=0; 16 for(int i=0; i<l ;i++) 17 { 18 19 for(int j=0; j<n; j++) 20 { 21 22 sum[cas++]=str1[i]+str2[j]; 23 } 24 } 25 return cas; 26 } 27 28 bool BSearch(long long sum[],int k,int cas) 29 { 30 31 int left=0,right=cas-1; 32 while(left<=right) 33 { 34 35 int mid=(left+right)>>1; 36 if(sum[mid] == k) return true; 37 else if(k < sum[mid]) right=mid-1; 38 else left = mid+1; 39 } 40 return false; 41 } 42 43 int main() 44 { 45 46 int l,m,n,i,j; 47 int ss=1; 48 while(scanf("%d%d%d",&l,&n,&m)!=EOF) 49 { 50 51 memset(str1,0,sizeof(str1)); 52 memset(str2,0,sizeof(str2)); 53 memset(str3,0,sizeof(str3)); 54 55 for(i=0; i<l; i++) 56 scanf("%d",&str1[i]); 57 for(j=0; j<n; j++) 58 scanf("%d",&str2[j]); 59 for(i=0; i<m; i++) 60 scanf("%d",&str3[i]); 61 int k,tmp; 62 scanf("%d",&k); 63 int cas=Table(l,n,m); 64 sort(sum,sum+cas); 65 printf("Case %d:\n",ss++); 66 while(k--) 67 { 68 scanf("%d",&tmp); 69 for(i=0; i<m; i++) 70 { 71 72 if(BSearch(sum,tmp-str3[i],cas)) 73 { 74 75 printf("YES\n"); 76 break; 77 } 78 } 79 if(i>=m)printf("NO\n"); 80 } 81 82 } 83 return 0; 84 }