HDU - 2141 Can you find it?
题目描述：
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
题目大意：就是给你三个序列，然后给你一个数，在三个序列中各找一个数，使得找的这三个数等于所给数，找得到输出“YES”，不然输出“NO”
思路：暴力当然能找得到，但是时间复杂度就是LMNS，毋庸置疑。。。超时的
找数嘛，想到二分，就是先在把两个序列合成一个序列，然后再查找，这样复杂度就降下来了！记得先sort一下
不多说，上代码！！！
#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
using namespace std;
int l , n , m , s , x;
int a[510];
int b[510];
int c[510];
int sum[250010];
int main(){
int w = 1;
while(~scanf("%d%d%d",&l,&n,&m)){
for(int i = 0 ; i < l ; ++i){
scanf("%d",&a[i]);
}
for(int i = 0 ; i < n ; ++i){
scanf("%d",&b[i]);
}
int k = 0;
for(int i = 0 ; i < l ; ++i){
for(int j = 0 ; j < n;++j){
sum[k++] = a[i] + b[j];
}
}
sort(sum,sum+k);
for(int i = 0 ; i < m ; ++i){
scanf("%d",&c[i]);
}
sort(c,c+m);
scanf("%d",&s);
printf("Case %d:\n",w);
w++;
for(int i = 0 ; i < s ; ++i){
scanf("%d",&x);
bool flag = false;
for(int j = 0; j < m ;++j){
int temp = x - c[j];
int low = 0 , high = k - 1;
int mid;
while(low <= high){
mid = (low + high) / 2;
if(sum[mid] > temp){
high = mid-1;
}
else if(sum[mid] < temp){
low = mid+1;
}
else{
flag = true;
break;
}
}
if(flag){
printf("YES\n");
break;
}
}
if(!flag) printf("NO\n");
}
}
return 0;
}