UVA 10771 Barbarian tribes(思维 | 线段树模拟约瑟夫环)

题意:

$1\le N + M\le 2000,1\le K\le 1000,N+M个人围成环,前N为G,后M为K$
$现在每轮:$
$每K个各杀1个,杀2个,添加一个到第2个死的位置上,相同加G,不同加K$
$也就是说每轮死1个,N+M-1轮后只剩1个,问是G还是K$

分析:

$线段树求K-th number的姿势直接暴力模拟$
$需要注意的是,1个死了,当前名次要-1,1个没死不动就好$
$时间复杂度为O((N+M)log(N+M)),跑了1.6s$
$也就是说这题起码有1W组case。。。$
$你特么不说一声,让我浪费时间写链表模拟。。O((N+M)K)直接T成sb了$


$其实有O(1)解法,设G为0,K为1,GG和KK变G,GK和KG变K$
$这其实就是异或,然后异或满足交换律,显然有奇数个1答案才是K$
$所以一行代码:$

1
puts(M & 1 ? "Keka" : "Gared");

线段树模拟代码:

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//
// Created by TaoSama on 2016-03-26
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, m, k;
int val[N << 2], sum[N << 2];
void build(int l, int r, int rt) {
sum[rt] = r - l + 1;
if(l == r) {
val[rt] = l <= n ? 1 : 2;
return;
}
int m = l + r >> 1;
build(l, m, rt << 1);
build(m + 1, r, rt << 1 | 1);
}
pair<int, int> query(int k, int l, int r, int rt) {
--sum[rt];
if(l == r) return make_pair(l, val[rt]);
int m = l + r >> 1;
if(sum[rt << 1] >= k) return query(k, l, m, rt << 1);
return query(k - sum[rt << 1], m + 1, r, rt << 1 | 1);
}
void update(int o, int v, int l, int r, int rt) {
if(l == r) {
val[rt] = v;
sum[rt] = 1;
return;
}
int m = l + r >> 1;
if(o <= m) update(o, v, l, m, rt << 1);
else update(o, v, m + 1, r, rt << 1 | 1);
sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d%d%d", &n, &m, &k) == 3 && (n || m || k)) {
build(1, n + m, 1);
n += m;
int rk = k;
for(int i = 1; i <= n - 1; ++i) {
int lft = n - i;
int last = query(rk, 1, n, 1).second;
rk = (rk - 1 + k) % lft;
if(rk == 0) rk = lft;
pair<int, int> now = query(rk, 1, n, 1);
int v = last == now.second ? 1 : 2;
update(now.first, v, 1, n, 1);
rk = (rk + k) % lft;
if(rk == 0) rk = lft;
}
int ans = query(1, 1, n, 1).second;
puts(ans == 1 ? "Gared" : "Keka");
}
return 0;
}


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