test20181006 石头剪刀布

题意

分析

考场做法同题解一样。

std代码。

#include<bits/stdc++.h>
using namespace std;

template <typename T> void chmin(T&x,const T &y)
{
	if(x>y)x=y;
}
template <typename T> void chmax(T &x,const T &y)
{
	if(x<y)x=y;
}
typedef long long s64;
typedef unsigned long long u64;
typedef unsigned int u32;
typedef pair<int,int> pii;
#define rep(i,l,r) for(int i=l;i<=r;++i)
#define per(i,r,l) for(int i=r;i>=l;--i)
#define rep0(i,l,r) for(int i=l;i<r;++i)
#define gc (c=getchar())
int read()
{
	char c;
	while(gc<'-');
	if(c=='-')
	{
		int x=gc-'0';
		while(gc>='0')x=x*10+c-'0';
		return -x;
	}
	int x=c-'0';
	while(gc>='0')x=x*10+c-'0';
	return x;
}
#undef gc

int n,need_a[3];
const char s[]="RPS";
string write(int n,int x)
{
	if(!n)
	{
		string ans;
		ans.push_back(s[x]);
		return ans;
	}
	string a=write(n-1,x),b=write(n-1,(x+1)%3);
	return a<b?a+b:b+a;
}
bool check(int x)
{
	int a[3]={},a0[3];
	a[x]=1;
	rep(tmp,1,n)
	{
		rep(i,0,2)a0[i]=a[i];
		rep(i,0,2)a[(i+1)%3]+=a0[i];
	}
	//cerr<<a[0]<<" "<<a[1]<<" "<<a[2]<<endl;
	rep(i,0,2)
	if(a[i]!=need_a[i])return 0;
	cout<<write(n,x)<<endl;
	return 1;
}

int main()
{
	freopen("rps.in","r",stdin);freopen("rps.out","w",stdout);
		rep(i,0,2)need_a[i]=read();
		n=0;
		while((1<<n)!=need_a[0]+need_a[1]+need_a[2])++n;
		int i=0;
		for(;i<=2;++i)
		if(check(i))break;
		if(i>2)puts("IMPOSSIBLE");
}

我的代码。排序的时候我用的递推处理，跑得比std快。

#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<ctime>
#include<iostream>
#include<string>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<complex>
#pragma GCC optimize ("O0")
using namespace std;
template<class T> inline T read(T&x)
{
    T data=0;
	int w=1;
    char ch=getchar();
    while(!isdigit(ch))
    {
		if(ch=='-')
			w=-1;
		ch=getchar();
	}
    while(isdigit(ch))
        data=10*data+ch-'0',ch=getchar();
    return x=data*w;
}
typedef long long ll;
const int INF=0x7fffffff;

const int MAXN=1<<20|7;

int a[2][MAXN],len[2];
int cur;

int main()
{
  freopen("rps.in","r",stdin);
  freopen("rps.out","w",stdout);
	int R,P,S;
	read(R);read(P);read(S);
	int n;
	for(int i=1;i<=20;++i)
		if(R+P+S==(1<<i))
			n=i;
//	cerr<<"n="<<n<<endl;
	for(int i=1;i<=3;++i)
	{
		cur=0;
		len[cur]=0;
		a[cur][++len[cur]]=i;
		for(int j=1;j<=n;++j)
		{
			cur^=1;
			len[cur]=0;
			for(int k=1;k<=len[cur^1];++k)
			{
				if(a[cur^1][k]==1)
				{
					a[cur][++len[cur]]=1;
					a[cur][++len[cur]]=2;
				}
				else if(a[cur^1][k]==2)
				{
					a[cur][++len[cur]]=2;
					a[cur][++len[cur]]=3;
				}
				else
				{
					a[cur][++len[cur]]=1;
					a[cur][++len[cur]]=3;
				}
			}
		}
		int r=0,p=0,s=0;
		for(int j=1;j<=len[cur];++j)
		{
			if(a[cur][j]==1)
				++p;
			else if(a[cur][j]==2)
				++r;
			else
				++s;
		}
		if(R==r&&P==p&&S==s)
		{
			for(int l=1;l<=n;++l) // the log of the sum of swap len
			{
				int d=(1<<l); // the sum of swap len
				for(int j=1;j<len[cur];j+=d)
				{
					int x=j,y=j+(d>>1);
					while(a[cur][x]==a[cur][y]&&x<j+(d<<1))
						++x,++y;
					if(x<j+(d<<1))
					{
//						cerr<<"x="<<x<<" y="<<y<<endl;
//						cerr<<"cmp"<<a[cur][x]<<" "<<a[cur][y]<<endl;
						if(a[cur][x]<a[cur][y])
							continue;
//						cerr<<"swap fr "<<j<<" to "<<(j+(d>>1))<<endl;
						for(int k=j;k<j+(d>>1);++k)
							swap(a[cur][k],a[cur][k+(d>>1)]);
					}
				}
			}
			for(int j=1;j<=len[cur];++j)
			{
				if(a[cur][j]==1)
					putchar('P');
				else if(a[cur][j]==2)
					putchar('R');
				else
					putchar('S');

			}
			puts("");
			return 0;
		}
	}
	puts("IMPOSSIBLE");
//  fclose(stdin);
//  fclose(stdout);
    return 0;
}