题目连接：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): 18192 Accepted Submission(s): 4601
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
题意： 从三个集合中分别各取一个数，然后相加等于X
注意数据范围很大所以不可能暴力，先算出前两个集合中任意两个值得和保存到数组d中，然后对d进行去重操作，注意：去重函数unique(d,d+cc)返回的是最后一个元素所在的地址，要获得新数组的总元素个数要减去首地址及 int n = unique(d,d+cc) - d ;
这里要注意，两个数相加可能会超int，所以要用long long，而且在每次输入x的时候在对应的d数组中找是否存在x-c的时候必须用O(n)的算法，所以考虑到用哈希的方法，最慢的哈希也要比二分快。
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 #define ll long long 5 ll a[505],b[505],c[505],d[250010]; 6 7 const int Mod = 500007; 8 int inf = 100000000; 9 ll mp[Mod]; 10 void insert(ll u) 11 { 12 ll v = u; 13 if(v<0) v*=-1; 14 int id = v%Mod; 15 while(mp[id]!=inf) {id++;id%=Mod;} 16 mp[id] = u; 17 } 18 bool find(ll u){ 19 ll v = u; 20 if(v<0) v*=-1; 21 int id = v%Mod; 22 while(mp[id] != inf){ 23 if(mp[id]==u) return true; 24 id++; 25 id%=Mod; 26 } 27 return false; 28 } 29 int main() 30 { 31 inf = inf*inf;//这里超int也没有关系，因为开始定义的int不够大 32 int n1 , n2 , n3 ,cas = 1; 33 while(~scanf("%d%d%d",&n1,&n2,&n3)) 34 { 35 for(int i =0 ; i < n1; i++) scanf("%lld",&a[i]); 36 for(int i =0 ;i < n2 ; i++) scanf("%lld",&b[i]); 37 for(int i =0 ;i < n3; i++) scanf("%lld",&c[i]); 38 int cc = 0; 39 for(int i =0 ;i < n1; i++) 40 for(int j = 0 ; j < n2 ; j++) 41 d[cc++] = a[i]+b[j]; 42 sort(d,d+cc); 43 int n = unique(d,d+cc)-d; 44 for(int i = 0 ; i < Mod ;i++) mp[i] = inf; 45 for(int i =0 ;i < n; i++) insert(d[i]); 46 ll s; 47 scanf("%lld",&s); 48 printf("Case %d: ",cas++); 49 for(int i = 0 ;i < s ;i++){ 50 ll x; 51 scanf("%lld",&x); 52 bool flag = false; 53 for(int j= 0 ; flag == false&&j<n3;j++) 54 if(find(x-c[j]))flag = true; 55 if(flag) puts("YES"); 56 else puts("NO"); 57 } 58 } 59 return 0; 60 }