HDU 2141 二分查找，实现上简单，思想很重要

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

const int N = 505;

__int64 ab[N * N];
int num;

int search(__int64 x)
{
int f = 0, l = num - 1;
int mid;
while(f <= l)
{
mid = (f + l) / 2;
if(ab[mid] == x)
return 1;
else if(ab[mid] < x)
f = mid + 1;
else
l = mid - 1;
}
return 0;
}

int main()
{
int n, m, l, flag = 0, s;
__int64 a
, b
, c
, x;
while(cin >> n >> m >> l)
{
flag++;
num = 0;
for(int i = 0; i < n; i++)
scanf("%I64d", &a[i]);
for(int i = 0; i < m; i++)
scanf("%I64d", &b[i]);
for(int i = 0; i < l; i++)
scanf("%I64d", &c[i]);

for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
ab[num++] = a[i] + b[j];
sort(ab, ab+num);
sort(c, c+l);
scanf("%d", &s);
printf("Case %d:\n", flag);
while(s--)
{
scanf("%I64d", &x);
if(x < ab[0] + c[0] || x > ab[num-1] + c[l-1])
printf("NO\n");
else
{
__int64 p;
int j;
for(j = 0; j < l; j++)
{
p = x - c[j];
if(search(p))
{
printf("YES\n");
break;
}
}
if(j == l)
printf("NO\n");
}
}
}
return 0;
}