HDU/5499/模拟

题目链接
模拟题,直接看代码。
£:分数的计算方法,要用double;

#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int, int> pa;
typedef long long  LL;
const int maxn=100+100;
struct node
{
    string na;
    double a;
    double b;
    double ans;
    bool operator<(const node x)const
    {
        return x.ans<ans;
    }
};
node man  [maxn];
node woman[maxn];
node t    [maxn];
string x1,x2;
double t1=0.0,t2=0.0;
void init()
{
    for(int i=0;i<=100+1;i++)
    {
        woman[i].ans=0;
        man  [i].ans=0;
        t    [i].ans=0;
    }
}

int main ()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        int n,m;
        scanf("%d%d",&n,&m);
        int k1=0,k2=0;
        bool fg=0;
        double xx=0.0,yy=0.0;
        for(int i=0;i<n;i++)
        {
            cin>>x1>>x2>>t1>>t2;
            if(x2[0]=='m')
            {
                man[k1].na=x1;
                man[k1].a=t1;
                man[k1].b=t2;
                xx=max(xx,t1);
                yy=max(yy,t2);
                k1++;
            }
            else
            {
                fg=1;
                woman[k2].na=x1;
                woman[k2].a=t1;
                woman[k2].b=t2;
                xx=max(xx,t1);
                yy=max(yy,t2);
                k2++;
            }
        }



        xx=(double)300.0/(double)xx;
        yy=(double)300.0/(double)yy;
        for(int i=0;i<k1;i++)
            man[i].ans=(double)man[i].a*xx*0.3+man[i].b*yy*0.7;
        for(int i=0;i<k2;i++)
            woman[i].ans=(double)woman[i].a*xx*0.3+woman[i].b*yy*0.7;


    sort(woman,woman+k2);




        if(fg)
        {
            int k=0;
            t[k].na=woman[0].na;
            t[k].ans=woman[0].ans;
            k++;
            for(int i=1;i<k2;i++)
            {
                man[k1].na=woman[i].na;
                man[k1].ans=woman[i].ans;
                k1++;
            }
            sort(man,man+k1);
            for(int i=0;i<m-1;i++)
            {
                t[k].na=man[i].na;
                t[k].ans=man[i].ans;
                k++;
            }
            sort(t,t+m);
            cout<<"The member list of Shandong team is as follows:"<<endl;
            for(int i=0;i<m;i++)
                cout<<t[i].na<<endl;
        }
        else
        {
            for(int i=0;i<k2;i++)
            {
                man[k1].na=woman[i].na;
                man[k1].ans=woman[i].ans;
                k1++;
            }
            sort(man,man+k1);
            cout<<"The member list of Shandong team is as follows:"<<endl;
            for(int i=0;i<m;i++)
                cout<<man[i].na<<endl;
        }
    }
    return 0;
}
想的太多,做的太少。
原文地址:https://www.cnblogs.com/pealicx/p/6115611.html