HDOJ 2141 Can you find it?

Can you find it?

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

题目大意：给出3个数组，问能否从3个数组中个找一个数，使得它们的和为X。

解题思路：利用数组sum存储a[i]与b[j]的和，然后对数组sum排序，二分搜索sum数组，找出满足sum[mid] == x - c[i]的mid，设置flag = true，打印输出即可。

代码如下：

#include <cstdio>
#include <algorithm>
typedef long long ll;
using namespace std;
const int maxl = 505;
int la,lb,lc,a[maxl],b[maxl],c[maxl],sum[maxl * maxl];
int k;
bool bs(int t)
{
int low,up,mid;
low = 0,up = k - 1;
while(low <= up){
mid = (low + up) / 2;
if(sum[mid] > t)
up = mid - 1;
else if(sum[mid] < t)
low = mid + 1;
else
return true;
}
return false;
}

int main()
{
int i,j,ncase = 0,s;
while(scanf("%d %d %d",&la,&lb,&lc) != EOF){
for(i = 0;i < la;i++)
scanf("%d",&a[i]);
for(i = 0;i < lb;i++)
scanf("%d",&b[i]);
for(i = 0;i < lc;i++)
scanf("%d",&c[i]);
k = 0;
for(i = 0;i < la;i++){
for(j = 0;j < lb;j++){
sum[k] = a[i] + b[j];
k++;
}
}
sort(sum,sum + k);

scanf("%d",&s);
printf("Case %d:\n",++ncase);
while(s--){
int x;
scanf("%d",&x);
bool flag = false;
for(i = 0;i < lc;i++){
if(bs(x - c[i])){
flag = true;
break;
}
}
if(!flag)
printf("NO\n");
else
printf("YES\n");
}
}
return 0;
}