/*
唉，那個第二種策略應該是沒用吧。
為什麽我列出式子最優策略跟第二種策略有關？？
為什麽我寫了寫分布列還是跟第二種策略有關？？
唉。人生啊。 
*/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <iostream>

using namespace std;

const int maxn=1000+15;
int n;
double p[maxn][maxn];
double
 
 f[maxn][maxn];
int main()
{
    freopen("game.in","r",stdin);
    freopen("game.out","w",stdout);
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
     for (int j=0;j<i;j++)
      scanf("%lf",&p[i][j]);
    f[0][0]=1;
    for (int i=0;i<n;i++)
     for (int j=0;j<=i;j++)
      {
          f[i
 
+1][j+1]+=f[i][j]*p[i+1][j];
        f[i+1][j]+=f[i][j]*(1-p[i+1][j]); 
      }
    double ans=0;
    for (int j=0;j<=n;j++) ans+=j*f[n][j];
    printf("%.2lf\n",ans);
    return 0;
}

/*
一道比較好寫的折半搜索題。
估計暴搜剪枝也能差不多A。
*/
#include<iostream>
#include<cstdio>
#include
 
<cstring>
#include<algorithm>

#define N 50
#define inf 0x3f3f3f3f

using namespace std;
int n,m,ans,cnt1,cnt2,dep,flag;
int v[N],sum1[10485780],sum2[10485780];

bool cmp(int a,int b){return a>b;}

inline int read()
{
    int x=0,f=1;char c=getchar();
    while(c>‘9‘||c<‘0‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}

void dfs(int k,int rest)
{
    if(k==dep)
    {
        if(flag)
        {
            sum1[++cnt1]=rest;return;
        }
        else
        {
            sum2[++cnt2]=rest;return;
        }
    }
    dfs(k+1,rest);
    dfs(k+1,rest-v[k+1]);
}

int main()
{
    freopen("cake.in","r",stdin);
    freopen("cake.out","w",stdout);
    n=read();m=read();
    for(int i=1;i<=n;i++) v[i]=read();    
    dep=n/2;flag=1;dfs(0,m/2);
    dep=n;flag=0;dfs(n/2,m-m/2);
    int L=1,R=1;ans=inf;
    sort(sum1+1,sum1+cnt1+1);sort(sum2+1,sum2+cnt2+1,cmp);
    if(sum1[1]+sum2[cnt2]>=0) ans=sum1[1]+sum2[cnt2];
    else
    {
        while(L<=cnt1 && R<=cnt2)
        {
            while(sum1[L]+sum2[R]<0) L++;
            ans=min(ans,sum1[L]+sum2[R]); 
            while(sum1[L]+sum2[R]>0)
            {
                R++;
                ans=min(ans,sum1[L]+sum2[R-1]);
            } 
            L++;
        }
    }
    printf("%d\n",ans);
    fclose(stdin);fclose(stdout);
    return 0;
}

/*
20dfs+特判n=1,2
下次再忘了取模剁手了。 
*/ 
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <iostream>

#define N 1001
#define mod 1000000007
#define ll long long

using namespace std;
ll n,m,h,w,ans,cnt;
ll a,b,c,d,v;
ll Mar[N][N];
struct node{
    ll a,b,c,d,v;
}ask[N];

inline ll read()
{
    ll x=0,f=1;char c=getchar();
    while(c>‘9‘||c<‘0‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}

ll ksm(ll a,ll b)
{
    ll res=1;
    while(b)
    {
        if(b&1) res=res*a%mod;
        b>>=1;a=a*a%mod;
    }return res;
}

void solve1()
{
    for(int i=1;i<=n;i++)
    {
        a=read();b=read();c=read();d=read();v=read();
        for(int x=a;x<=c;x++) for(int y=b;y>=d;y--)
          Mar[x][y]=v;
    }
    ans=1;
}

bool judge()
{
    ll mx=0;
    for(int i=1;i<=n;i++)
    {
        for(int x=ask[i].a;x<=ask[i].c;x++) for(int y=ask[i].d;y>=ask[i].b;y--)
        {
            if(Mar[x][y]>ask[i].v) return false;
            mx=max(mx,Mar[x][y]);
        }
        if(mx!=ask[i].v) return false;
    }return true;
}

void dfs(int x,int y)
{
    if(y>w) x++,y%=w;
    if(x<1 || y<1) return;
    if(x==h+1)
    {
        if(judge()) ans++;ans%=mod;
        return;
    }
    for(int i=1;i<=m;i++)
    {
        Mar[x][y]=i;
        dfs(x,y+1);
    }
    
}

int main()
{
    freopen("grid.in","r",stdin);
    freopen("grid.out","w",stdout);
    h=read();w=read();m=read();n=read();
    if(n==0)
    {
        cout<<ksm(m,h*w)%mod<<endl;
        return 0;
    }
    if(n==1)
    {
        a=read();b=read();c=read();d=read();v=read();
        ll num=(c-a+1)*(d-b+1);
        ll res1=ksm(m,h*w-num)%mod;
        ll res2=ksm(v,num)%mod;
        ll res3=(res1%mod*ksm(v-1,num)%mod)%mod;
        ans=(res1%mod*res2%mod-res3+mod)%mod;
        cout<<ans%mod<<endl;
        return 0;
    }
    if(h<=3 && w<=3) 
    {
        for(int i=1;i<=n;i++)
        {
            ask[i].a=read();ask[i].b=read();
            ask[i].c=read();ask[i].d=read();
            ask[i].v=read();
        }
        dfs(1,1),cout<<ans<<endl;
    }
    fclose(stdin);fclose(stdout);
    return 0;
}

11.4NOIP模擬賽