hdoj 2141 Can you find it?【二分查找+暴力】

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 17036 Accepted Submission(s):
4337


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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".

[align=left]Sample Input[/align]

3 3 3
1 2 3
1 2 3
1 2 3
3
1
4

10

[align=left]Sample Output[/align]

Case 1:
NO

YES
NO
题意：给你三组数，每组数的个数分别为L N M再给你S个数x判断能否从这三组数中分别找到一个数使Ai+Bj+Ck = X.
题解：暴力肯定超时，所以要用二分，先将三组数中任意两组存入数组str[]中，然后再对每个x进行遍历
1、测试数据太弱，而且这样的一组测试数据本来就是个坑，好多不正确的情况都能得出正确答案
2、输出格式好坑，本来我以为是连续输出NO YES NO一直错没想到竟然是输一个数出一个结果
3、一定要注意给str[]数组排序（二分查找的条件是有序数列）

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAX 1100
bool cmp(int a,int b)
{
return a<b;
}
int a[MAX];
int b[MAX];
int c[MAX];
int d,str[300000];
int main()
{
int n,m,j,i;
int k=1,l;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
for(i=0;i<l;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
scanf("%d",&b[i]);
for(i=0;i<m;i++)
scanf("%d",&c[i]);
int len=0;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
str[len++]=b[i]+c[j];
}
}
sort(str,str+len,cmp);
int ok;
printf("Case %d:\n",k++);
int s;
scanf("%d",&s);
for(i=0;i<s;i++)
{
scanf("%d",&d);
ok=0;
for(j=0;j<l;j++)
{
int left = 0,right = len,mid = 0;
int goal = d - a[j];
while(right >= left)
{
mid = (right + left) >> 1;
if(str[mid] < goal)
left = mid + 1;
else if(str[mid] > goal)
right = mid - 1;
else
{
ok=1;
break;
}
}
if(ok)
break;
}
if(ok)
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}