HDOJ 2141 Can you find it?

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 20106 Accepted Submission(s): 5099



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

用三个for循环会超时

后来就改进了一下把函数改为：A+B=X-C，然后二分搜一下就可以了；

#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 500 + 10;

long long ab[maxn * maxn], a[maxn], b[maxn], c[maxn];
long long len;
int binary_search(long long kk)
{
long long fir = 0, las = len - 1;
while (fir <= las){
long long middle = (fir + las) / 2;
if (ab[middle] == kk)
return 1;
else if (ab[middle] > kk)
las = middle - 1;
else
fir = middle + 1;
}
return 0;
}

int main()
{
long long l, n, m, s, x;
long long ans, flag;
ans = 0;
while (scanf("%I64d%I64d%I64d", &l, &n, &m) != EOF){
len = 0;
for (int i = 0; i < l; i++)
scanf("%I64d", &a[i]);
for (int i = 0; i < n; i++)
scanf("%I64d", &b[i]);
for (int i = 0; i < m; i++)
scanf("%I64d", &c[i]);

for (int i = 0; i < l; i++){
for (int j = 0; j < n; j++){
ab[len++] = a[i] + b[j];
}
}
sort (ab, ab + len);
sort (c, c + m);
scanf("%I64d", &s);
printf("Case %I64d:\n", ++ans);
while (s--){
scanf("%I64d", &x);
if (x < ab[0] + c[0] || x > ab[len - 1] + c[m - 1]){
printf("NO\n");
continue;
}
flag = 0;
for (int i = 0; i < m; i++){
long long kk = x - c[i];
if (binary_search(kk)){
flag = 1;
break;
}
}
if (flag)
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}