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hdu 2141 （二分）
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 }
hdu 2141 （二分）
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