• BestCoder #47 1001&&1002


    【比赛链接】clikc here~~

    ps:真是wuyu~~做了两小时。A出两道题,最后由于没加longlong所有被别人hack掉!,最后竟然不知道hack别人不成功也会掉分。还一个劲的hack 别人的代码,昨天真是个悲催的比赛,~~~~(>_<)~~~~,以下弱弱献上代码~~

    1002比1001还简单~~

    1002  Senior's Gun

    /*
    BestCoder Round #47
    1002   Senior's Gun
    
    */
    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    
    using namespace std;
    
    #define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)
    #define per(i,j,k) for(int i=(int)j;i>=(int)k;i--)
    typedef long long LL;
    typedef unsigned long long LLU;
    typedef double db;
    
    const int N =2*1e5+10;
    int n,m,t,p,res,cnt;
    LL ans,tmp;
    int num[N];
    int aa[N],bb[N];
    char str[N];
    bool vis[N];
    
    int main()
    {
        scanf("%d",&t);
        while (t--)
        {
            scanf("%d%d",&n,&m);
            for (int i=1; i<=n; i++) scanf("%d",&aa[i]);
            for (int i=1; i<=m; i++) scanf("%d",&bb[i]);
            sort(aa+1,aa+n+1);
            sort(bb+1,bb+m+1);
            ans=0;
            int j=n;
            for (int i=1; i<=min(n,m); i++)
                if(aa[j]>bb[i])
                {
                    ans+=aa[j]-bb[i];
                    j--;
                }
                else break;
            printf("%I64d
    ",ans);
        }
        return 0;
    }

    Sample Input
    1
    2 2
    2 3
    2 2
    
    Sample Output
    1

    1001 Senior's Array

    /*
    BestCoder Round #47
    1001   Senior's Array
    
    */
    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    
    using namespace std;
    
    #define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)
    #define per(i,j,k) for(int i=(int)j;i>=(int)k;i--)
    typedef long long LL;
    typedef unsigned long long LLU;
    typedef double db;
    
    const int N =2*1e4+10;
    int n,m,t,p;
    int aa[N],bb[N];
    char str[N];
    bool vis[N];
    
    int main()
    {
        scanf("%d",&t);
        while (t--)
        {
            scanf("%d%d",&n,&p);
            rep(i,1,n) scanf("%d",&aa[i]);
            LL ans=-1e9;
            rep(i,1,n)
            {
                int tmp=aa[i];
                aa[i]=p;
                LL now=0;
                rep(j,1,n)
                {
                    now+=(1ll)*aa[j];
                    if(now>ans) ans=now;
                    if(now<0) now=0;
                }
                aa[i]=tmp;
            }
            printf("%I64d
    ",ans);
        }
        return 0;
    }


    Sample Input
    2
    3 5
    1 -1 2
    3 -2
    1 -1 2
    Sample Output
    8
    2

  • 相关阅读:
    冒泡排序
    二分查找
    数3退1的java实现
    列出目录下对应的子目录及文件
    errno相关
    Fibonacci
    windows 下查看Python 的安装路径
    20条编程经验
    [转]一位程序员工作10年总结的13个忠告
    Sql查询语句过滤重复的数据
  • 原文地址:https://www.cnblogs.com/lxjshuju/p/6863657.html
Copyright © 2020-2023  润新知