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  • HNU11376:Golf Bot

    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.
    Next line has one integer: M, the number of holes in this course. Each of the following M lines has one integer, dj , the distance from Golf Bot to hole j.
    1<=N,M<=200 000
    1<=ki,dj<=200 000


    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,给出每一个洞的位置。问两杆之内,在仅仅能往前打的情况下。能进的有几种洞


    思路:

    数据非常大?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
    ",ans);
        }
    
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/cxchanpin/p/6803154.html
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