标签：hnu
| Problem description |
|
|
| Input |
|
The first line has one integer: N, the number of different distances the Golf Bot can shoot. Each of the following N lines has one integer, ki, the distance marked in position i of the knob. |
| Output |
|
You should output a single integer, the number of holes Golf Bot will be able to complete. Golf Bot cannot shoot over a hole on purpose and then shoot backwards. |
| Sample Input |
3 1 3 5 6 2 4 5 7 8 9 |
| Sample Output |
4 |
| Problem Source |
| HNU Contest |
题意：
打高尔夫，一球能打n种距离，有m个洞，给出每个洞的位置，问两杆之内，在只能往前打的情况下，能进的有几种洞
思路：
数据很大？但你要看有30S，明摆着告诉你暴力可行
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 200005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define rank rank1
const int mod = 1000000007;
int hsh[N*2];
int a[N],b[N];
int main()
{
int n,m,i,j,k;
while(~scanf("%d",&n))
{
MEM(hsh,0);
for(i = 0; i<n; i++)
{
scanf("%d",&a[i]);
hsh[a[i]] = 1;
}
sort(a,a+n);
for(i = 0; i<n; i++)
{
for(j = i; j<n; j++)
{
hsh[a[i]+a[j]] = 1;
}
}
int ans = 0;
scanf("%d",&m);
for(i = 0; i<m; i++)
{
scanf("%d",&b[i]);
if(hsh[b[i]])
ans++;
}
printf("%d\n",ans);
}
return 0;
}
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标签：hnu
原文地址：http://blog.csdn.net/libin56842/article/details/47403569
