BestCoder Round #7-A,B,C

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A:Little Pony and Permutation

直接暴力搜尋，複雜度O（n）

#include<stdio.h>#include<iostream>#include<stdlib.h>#include<string.h>#include<algorithm>#include<vector>#include<math.h>#include<queue>#include<stack>#include<map>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define maxn 110000int a[maxn];int vis[maxn];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)scanf("%d",&a[i]);        memset(vis,0,sizeof(vis));        for(int i=1;i<=n;i++)        {            if(vis[i])continue;            int x=i;            cout<<"("<<x;            vis[x]=1;            x=a[x];            while(!vis[x])            {                vis[x]=1;                cout<<" "<<x;                x=a[x];            }            cout<<")";        }        cout<<endl;    }    return 0;}

B：Little Pony and Alohomora Part I

我們可以發現，這是一個調和序列。

然後。。。就很那個啥了。。、

#include<stdio.h>#include<iostream>#include<stdlib.h>#include<string.h>#include<algorithm>#include<vector>#include<math.h>#include<queue>#include<stack>#include<map>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define maxn 1100000double f[maxn];int main(){    int n;    f[1]=1.0;    for(int i=2;i<maxn;i++)f[i]=f[i-1]+1.0/i;    while(~scanf("%d",&n))    {        if(n<maxn)printf("%.4lf\n",f[n]);        else printf("%.4lf\n",log(n)+0.5772156649);    }    return 0;}

C：Little Pony and Dice

dp[i]: 恰好走到i點的機率

dp[i]=dp[i-1]*(1/m)+(dp[i-1]-dp[i-m-1]*(1.0/m))

dp[i-1]*(1/m)代表從i-1這個點走到i點的機率。

(dp[i-1]-dp[i-m-1]*(1.0/m))所有走到i-1的點除了i-m-1點外，其他的點都能走到i，並且走到i-1和走到i的機率相同

如果相鄰的兩個i之間的dp[i]相差很小，那說明下一個dp[i]的變化也會很小。以後就都是這個定值了。所以直接輸出就好。

<pre name="code" class="cpp">#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<vector>#include<set>#include<string>#include<math.h>using namespace std;#define LL __int64#define maxn 703000#define eps 1e-13#define zero(x) (fabs(x)<eps?0:x)double dp[maxn];int main(){    int n,m;    while(~scanf("%d%d",&m,&n))    {        dp[0]=1.0;        dp[1]=1.0/m;        int i;        for(i=2;i<=n;i++)        {            dp[i]=dp[i-1]+dp[i-1]*1.0/m;            if(i>m)dp[i]=dp[i]-dp[i-m-1]*1.0/m;            if((zero(dp[i]-dp[i-1])==0))            {                printf("%.5lf\n",dp[i]);                break;            }        }        if(i>n)printf("%.5lf\n",dp[n]);    }    return 0;}

BestCoder Round #7-A,B,C