大意: n个人, 两个党派, 轮流投票, 两种操作(1)ban掉一个人 (2)投票, 每轮一个未被ban的人可以进行一次操作(1)或操作(2), 求最终哪个党派得票最多.

 

显然先ban人会更优, 所以维护两个set模拟. 不过好像可以有O(n)的做法?

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head


const int N = 1e6+10;
int n, a[N];
char s[N], b[2][2]{"R","D"};
set<int> q, g[2];

int main() {
	scanf("%d%s", &n, s);
	REP(i,0,n-1) { 
		a[i]=s[i]=='D';
		q.insert(i);
		g[a[i]].insert(i);
	}
	REP(i,0,1) if (g[!i].empty()) return puts(b[i]),0;
	while (q.size()) {
		int x = *q.begin(); q.erase(x);
		int id = a[x%n], now = x/n;
		auto t = g[!id].lower_bound(x%n);
		if (t==g[!id].end()) t=g[!id].begin(),++now;
		q.erase(now*n+*t);
		g[!id].erase(t);
		if (g[!id].empty()) return puts(b[id]),0;
		q.insert(x+n);
	}
}