传送
因为每个人是最优策略,所以他会选中奖概率最高的箱子。
要知道当前哪一个箱子中奖概率最高,就应该求出一个人抽完奖后这个箱子期望有几个信封获奖,即(l_i-frac{l_i}{t_i})个。
如果这道题不要输出分数的话,用单调队列模拟(k)次就好了。
加上分数的条件,就是把(l_i)看成(frac{a_i}{b_i}),则抽完后,就是(a_i'=a_i(t_i-1),b_i'=b_it_i)。所以单调队列里面我们维护(t_i,a_i,b_i)就可以了。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<assert.h>
#include<ctime>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
//const int maxn = ;
In ll read()
{
ll ans = 0;
char ch = getchar(), las = ' ';
while(!isdigit(ch)) las = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(las == '-') ans = -ans;
return ans;
}
In void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
freopen(".in", "r", stdin);
freopen(".out", "w", stdout);
#endif
}
int n, K;
struct Node
{
ll t, a, b;
In bool operator < (const Node& oth)const
{
return a * oth.b * oth.t < oth.a * b * t;
}
};
priority_queue<Node> q;
In ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
int main()
{
// MYFILE();
int T = read();
while(T--)
{
while(!q.empty()) q.pop();
n = read(), K = read();
for(int i = 1; i <= n; ++i)
{
Node tp; tp.t = read(), tp.a = read(), tp.b = 1;
q.push(tp);
}
for(int i = 1; i < K; ++i)
{
Node tp = q.top(); q.pop();
ll a = tp.a * (tp.t - 1), b = tp.b * tp.t, g = gcd(a, b);
q.push((Node){tp.t, a / g, b / g});
}
Node tp = q.top();
ll a = tp.a, b = tp.b * tp.t, g = gcd(a, b);
write(a / g), putchar('/'), write(b / g), enter;
}
return 0;
}