Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others) Total Submission(s): 32126 Accepted Submission(s): 8011
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
HDU 2007-11 Programming Contest
Recommend
威士忌
思路：暴力解复杂度o(LMNS), 大约10^11，所以会超时，采用分治思想，将A+B+C=X，转化为A+B=X-C；算出A+B的数组，从数组中二分查X-C；这样时间复杂度降为O（LN+MS*log（LN））， 约等于O（LN+MS）；就不会超时了；
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int maxn=505;
int a[maxn], b[maxn], c[maxn], x[2*maxn], ab[maxn*maxn];
int l, n, m, s, cnt1, cas=0;
bool binarySearch(int tgt)
{
int l=1, r=cnt1, mid;
while(r>=l)
{
mid=(l+r)>>1;
if(ab[mid]==tgt)
return true;
else if(ab[mid]>tgt)
r=mid-1;
else
l=mid+1;
}
return false;
}
int main(){
while(~scanf("%d%d%d", &l, &n, &m))
{
cnt1=0;
for(int i=1; i<=l; i++)
scanf("%d", a+i);
for(int i=1; i<=n; i++)
scanf("%d", b+i);
for(int i=1; i<=m; i++)
scanf("%d", c+i);
scanf("%d", &s);
for(int i=1; i<=s; i++)
scanf("%d", x+i);
printf("Case %d:\n", ++cas);
for(int i=1; i<=l; i++)
for(int j=1; j<=n; j++)
ab[++cnt1]=a[i]+b[j];
sort(ab+1, ab+1+cnt1);
for(int i=1; i<=s; i++)
{
int flag=1;
for(int j=1; j<=m; j++)
{
if(binarySearch(x[i]-c[j])==true)
{
printf("YES\n");
flag=0;
break;
}
}
if(flag)
printf("NO\n");
}
}
return 0;
}
