zjnu 1739 PŠENICA()

Description

Our heroes, Mirko and Slavko, plant Christmas wheat every year on Saint Lucy’s Day. It is well known

that stalks of wheat grow at different speeds so, after a certain time, the wheat becomes quite messy.

The guys are determined to solve this problem by playing the following game:

When it’s Mirko’s turn, he chooses a stalk of wheat with the minimal height and prolongs its

height so it’s of the same height as the first stalk longer than it.

When it’s Slavko’s turn, he chooses a stalk of wheat with the maximal height and cuts it to be

of the same height as the first stalk shorter than it.

The game lasts while there are at least three stalks of different heights and the loser is the player

who can’t make his move.

For given heights of all stalks of wheat and the assumption that Mirko is the one starting the game,

determine the winner of the game and the height of the shortest and longest stalk when the game is

finished.

Input

The first line of input contains the integer N (1 <= N <= 100000), the number of wheat stalks.

The second line of input contains N space separated integers that denote the heights of individual

wheat stalks. The height of each stalk will be less than or equal to 100000.

Output

The first line of output must contain the word “Mirko” if Mirko is the winner of the game, or “Slavko”

if Slavko is the winner of the game.

The second line of output must contain the height of the shortest and longest stalk when the game is

finished.

Sample Input

3
3 3 3
Sample Output

Slavko
3 3
题意就是给你一个序列，两个人玩游戏，第一个人要把最小的数变成倒数第二小的数，第二个人要把最大的数变成倒数第二大的数，直达这个序列还剩两个数，谁先不能走谁输，输出赢得人。

这是一道很好的模拟题，如果你直接暴力是超时的，所以直接求前缀和成段更新。思路很棒！

AC代码：

#include <stack>
using namespace std;
typedef __int64 ll;
const int maxn=100010;
ll s[maxn],dig[maxn],a[maxn];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
ll i,n,tot;
scanf("%I64d",&n);
for(i=1;i<=n;i++)
scanf("%I64d",&a[i]);
sort(a+1,a+n+1);
memset(s,0,sizeof(s));
tot=1;
for(i=1;i<=n;)
{
dig[tot]=a[i];
for(;i<=n;i++)
{
if(dig[tot]==a[i]) s[tot]++;
else break;
}
tot++;
}
for(i=2;i<tot;i++)
s[i]+=s[i-1];
ll w=tot-1,tsum=0,wsum=0,t=1;
int last=1;
while(w-t>1)
{
if(tsum+s[t]<=wsum+s[tot-1]-s[w-1])
{
last=0;
tsum+=s[t];
t++;
}
else
{
last=1;
wsum+=s[tot-1]-s[w-1];
w--;
}
}
if(!last) cout<<"Mirko"<<endl;
else cout<<"Slavko"<<endl;
cout<<dig[t]<<" "<<dig[w]<<endl;
return 0;
}