2019.08.15【NOIP提高组】模拟 A 组

JZOJ 6296 投票

题目

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分析

根据贪心从小到大排序后选择两边的同学必然是最优的，所以可以写一个dp方程，枚举左边同学求最大值

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代码

#include <cstdio>
#include <algorithm>
#define rr register
using namespace std;
double p[2001],f[2001][2001],dp[2001][2001],ans; int n,m;
inline void pro(double f[2001][2001]){
	f[0][0]=1;
	for (rr int i=1;i<=m;++i)
	for (rr int j=0;j<=i;++j){
		f[i][j]=f[i-1][j]*(1-p[i]);
		if (j) f[i][j]+=f[i-1][j-1]*p[i];
	}
}
signed main(){
	freopen("vote.in","r",stdin);
	freopen("vote.out","w",stdout);
	scanf("%d%d",&n,&m);
	for (rr int i=1;i<=n;++i) scanf("%lf",&p[i]);
	sort(p+1,p+1+n),pro(f),reverse(p+1,p+1+n),pro(dp);
	for (rr int i=0;i<=m;++i){
		rr double t=0;
		for (rr int j=0;j<=m/2;++j)
		    t+=f[i][j]*dp[m-i][m/2-j];
		ans=ans>t?ans:t;
	}
	return !printf("%lf",ans);
} 

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JZOJ 6294 动态数点

题目

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分析

用ST表维护区间最小值和区间gcd，显然只有两值相等才可能为答案，二分答案，那么时间复杂度$O(nlogn)$，还没有$O(n^2)$的暴力快

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代码

#include <cstdio>
#include <cctype>
#include <cstring>
#define rr register
using namespace std;
const int N=500101;
typedef unsigned uit;
uit fgcd[N][20],mmin[N][20];
int n,q[N],q1[N],lg[N];
inline uit iut(){
	rr uit ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
inline uit gcd(uit x,uit y){return y?gcd(y,x%y):x;} 
inline uit min(uit x,uit y){return x<y?x:y;}
inline uit ggcd(int l,int r){
	rr int z=lg[r-l+1];
	return gcd(fgcd[l][z],fgcd[r-(1<<z)+1][z]); 
} 
inline uit gmin(int l,int r){
	rr int z=lg[r-l+1];
	return min(mmin[l][z],mmin[r-(1<<z)+1][z]);
}
inline bool check(int len){
	rr bool flag=0; q[0]=0;
	for (rr int i=1;i+len<=n;++i){
		if (ggcd(i,i+len)==gmin(i,i+len))
			flag=1,q[++q[0]]=i;
	}
	return flag;
}
signed main(){
	freopen("point.in","r",stdin);
	freopen("point.out","w",stdout);
	n=iut(); lg[0]=-1;
	for (rr int i=1;i<=n;++i) lg[i]=lg[i>>1]+1;
	for (rr int i=1;i<=n;++i) fgcd[i][0]=mmin[i][0]=iut();
	for (rr int j=1;j<=lg[n];++j)
	for (rr int i=1;i+(1<<j)-1<=n;++i){
		fgcd[i][j]=gcd(fgcd[i][j-1],fgcd[i+(1<<(j-1))][j-1]),
		mmin[i][j]=min(mmin[i][j-1],mmin[i+(1<<(j-1))][j-1]);
	}
	rr int l=0,r=n-1;
	while (l<r){
		rr int mid=(l+r+1)>>1;
		if (check(mid)) l=mid,memcpy(q1,q,sizeof(q1));
		    else r=mid-1;
	}
	printf("%d %d",q1[0],l);
	for (rr int i=1;i<=q1[0];++i) printf("%c%d",i==1?10:32,q1[i]);
	return 0;
}

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JZOJ 6303 演员

题目

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分析

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代码

//f表示dp方程，dp表示f的前缀和，fac和inv显然是阶乘和逆元
#include <cstdio>
#include <cctype>
#define rr register
using namespace std;
const int mod=1e9+7; bool now;
int n,m,fac[1000011],inv[1000011],ans,f[2][311][311],dp[2][311][311];
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
inline signed mo(int x,int y){return x+y>=mod?x+y-mod:x+y;}
signed main(){
	freopen("actor.in","r",stdin);
	freopen("actor.out","w",stdout);
	scanf("%d%d",&n,&m);
	inv[0]=inv[1]=fac[0]=fac[1]=f[0][0][0]=1,now=0;
	for (rr int i=2;i<=n;++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
	for (rr int i=2;i<=n;++i) fac[i]=1ll*fac[i-1]*i%mod,inv[i]=1ll*inv[i-1]*inv[i]%mod;
	for (rr int i=1;i<2*m;++i){
		if (i>n) break; f[now^=1][0][0]=dp[now][0][0]=0;
		for (rr int j=1;j<=m;++j){
			if (i<j) break;
			for (rr int k=1;k<=j;++k){
				f[now][j][k]=mo(1ll*f[now^1][j-1][k-1]*(m-j+1)%mod,mo(dp[now^1][j][m]-dp[now^1][j][k],mod)),//显然只有（m-j+1）种颜色可选
				
				dp[now][j][k]=mo(dp[now][j][k-1],f[now][j][k]),ans=mo(ans,1ll*f[now][j][k]*fac[n-1]%mod*inv[i-1]%mod*inv[n-i]%mod);//要把相同颜色的填回去，所以要乘上C(n-1,i-1)
		    }
			for (rr int k=j+1;k<=m;++k) dp[now][j][k]=dp[now][j][k-1];
		}
	}
	return !printf("%d",ans);
}