題目：

思路：針對四種排序方法構建四個結構體，按四種排序排完之後，把結果彙總到代表國家的一個結構體中。然後就是查詢就是了。排序規則可通過下面的例子瞭解一下：

序列：g[0] = 1,g[1] = 2，g[2] = 2, g[3] = 3;

排名：1          ,2        ,2          ,3(並不是1,2,3,4)

程式碼： 

#include <bits/stdc++.h>
#define FRE() freopen("in.txt","r",stdin)
using namespace std;
const int maxn = 500;
typedef unsigned long long ll;
int n,q;

struct Country
{
    int g,s;
    int g_p,s_p;
} ct[maxn];

struct node
{
    double gold_p, silver_p;
    int g, s;
    int id;
} g[maxn],s[maxn],g_p[maxn],s_p[maxn];

void init()
{
    cin>>n>>q;
    for(int i = 0; i<n; i++)
    {
        int num;
        cin>>g[i].g>>s[i].s>>num;
        g_p[i].gold_p = 1.0*g[i].g/num;
        s_p[i].silver_p = 1.0*s[i].s/num;
        g[i].id = s[i].id = g_p[i].id = s_p[i].id = i;
    }
    return;
}

bool cmdg(node& a, node& b)
{
    if(a.g == b.g)
        return a.id<b.id;
    return a.g > b.g;
}
bool cmds(node& a,node& b)
{
    if(a.s==b.s)
        return a.id<b.id;
    return a.s > b.s;
}
bool cmdg_p(node& a,node& b)
{
    return a.gold_p > b.gold_p;
}
bool cmds_p(node& a,node& b)
{
    return a.silver_p > b.silver_p;
}

int main()
{
    init();
    sort(g,g+n,cmdg);
    int lastid = 0;
    ct[g[0].id].g = 0;
    for(int i = 1; i<n; i++)
    {
        if(g[i].g==g[i-1].g)
            ct[g[i].id].g = lastid;
        else
        {
            ct[g[i].id].g = i;
            lastid = i;
        }
    }
    sort(s,s+n,cmds);
    lastid = 0;
    ct[s[0].s].s = 0;
    for(int i = 1; i<n; i++)
    {
        if(s[i].s==s[i-1].s)
            ct[s[i].id].s = lastid;
        else
        {
            ct[s[i].id].s = i;
            lastid = i;
        }
    }
    sort(g_p,g_p+n,cmdg_p);
    lastid = 0;
    ct[g_p[0].id].g_p = 0;
    for(int i = 1; i<n; i++)
    {
        if(g_p[i].gold_p == g_p[i-1].gold_p)
            ct[g_p[i].id].g_p = lastid;
        else
        {
            ct[g_p[i].id].g_p = i;
            lastid = i;
        }
    }
    sort(s_p,s_p+n,cmds_p);
    lastid = 0;
    ct[s_p[0].id].s_p = 0;
    for(int i = 1; i<n; i++)
    {
        if(s_p[i].silver_p == s_p[i-1].silver_p)
            ct[s_p[i].id].s_p = lastid;
        else
        {
            ct[s_p[i].id].s_p = i;
            lastid = i;
        }
    }
    for(int i = 0; i<q; i++)
    {
        int tmp,f,ans;
        cin>>tmp;
        f = 1,ans = ct[tmp].g;
        if(ct[tmp].s<ans)
            f = 2,ans = ct[tmp].s;
        if(ct[tmp].g_p<ans)
            f = 3,ans = ct[tmp].g_p;
        if(ct[tmp].s_p<ans)
            f = 4,ans = ct[tmp].s_p;
        if(i)
            cout<<" ";
        cout<<ans+1<<":"<<f;
    }
    return 0;
}
/*
樣例輸入：
4 4
51 100 1000
36 110 300
6 14 32
5 18 40
0 1 2 3
樣例輸出：
1:1 1:2 1:3 1:4
*/