2018 湘潭邀请赛 F.Sorting

思路一

long double 精度可能不够，直接交叉相乘需要 6 × (2 × 10 9 ) 2 > 2 64 .

转化为排序
c i
a i +b i +c i ，交叉相乘只需要

3 × (2 × 10 9 ) 2 < 2 64

注意：记得一定是：long double。交叉相乘也还是会炸，所以是对c/(a+b+c)排序;

#include <iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
struct node
{
    int id;
    long double v;
}w[1005];
bool cmp(node x,node y)
{
    if(x.v==y.v)
        return x.id<y.id;
    return x.v>y.v;
}
int main ()
{
    int n;
    while(cin>>n)
    {
        for (int i=0;i<n;i++)
        {
            long double a,b,c;
            cin>>a>>b>>c;
            w[i].v=c/(a+b+c);
            w[i].id=i+1;
        }
        sort(w,w+n,cmp);
        for (int i=0;i<n;i++)
            i==n-1?cout<<w[i].id<<endl:cout<<w[i].id<<" ";
    }
    return 0;
}

思路二：

#include <iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
struct node
{
    int id;
    long long a,b,c;
}w[1005];
bool cmp(node x,node y)
{
    return (x.a+x.b)*y.c<(y.a+y.b)*x.c;
}
int main ()
{
    int n;
    while(cin>>n)
    {
        for (int i=0;i<n;i++)
        {
            cin>>w[i].a>>w[i].b>>w[i].c;
            w[i].id=i+1;
        }
        stable_sort(w,w+n,cmp);
        for (int i=0;i<n;i++)
            i==n-1?cout<<w[i].id<<endl:cout<<w[i].id<<" ";
    }
    return 0;
}