概率专题·期望

uva 11021 Tribbles

k只麻球，每活一天就会死亡，但第二天可能会生一些麻球，具体说来，生i个麻球的概率为pi ,求m天后所有麻球都死亡的概率
fi=p0+p1fi−1+p2fi−22+...+pn−1fi−1n−1

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<iomanip> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (1000+10)int T,n,k,m;double p[MAXN],f[MAXN];int main(){//  freopen("uva11021.in","r",stdin);//  freopen(".out","w",stdout);    T=read();    For(kcase,T) {        n=read(),k=read(),m=read();        Rep(i,n) cin>>p[i];        f[0]=0; f[1]=p[0];        Fork(i,2,m) {            f[i]=0;            Rep(j,n) f[i]+=p[j]*pow(f[i-1],j);                              }        printf("Case #%d: %.7lf\n",kcase,pow(f[m],k));          }    return 0;}

UVA 11722 Joining with Friend

题意：假设你会在时间区间[t1,t2]中的任意时刻以相同的概率到达火车站，你朋友则在[s1,s2]中的任意时刻以相同的概率到达火车站,你们在到达后停留w分钟，问你们会面的概率。若存在一个时刻你们同时在火车站视为一定会面。
题解：概率空间是坐标系中的一个矩形，分情况讨论算面积。

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} ld t1,t2,s1,s2,w;int main(){//  freopen("uva11722.in","r",stdin);//  freopen(".out","w",stdout);    int T=read();    For(kcase,T) {        cin>>t1>>t2>>s1>>s2>>w;        if (s2-t2>s1-t1) swap(t1,s1),swap(t2,s2);        ld a[5]={0,s1-t2,s2-t2,s1-t1,s2-t1};        ld S=(t2-t1)*(s2-s1),p=0;        if (-w<a[1]) {        } else if (-w<a[2]) {            p+=pow(t2-w-s1,2)/2;            } else if (-w<a[3]) {            p+=(t2-s1-w+t2-s2-w)*(s2-s1)/2;        } else if (-w<a[4]) {            p+=S-pow(s2+w-t1,2)/2;        } else p+=S;        if (w<a[1]) {            p+=S;        } else if (w<a[2]) {            p+=S-(t2+w-s1)*(t2+w-s1) /2;            } else if (w<a[3]) {            p+=S- (t2-s1+w+t2-s2+w)*(s2-s1)/2;        } else if (w<a[4]) {            p+=pow(s2-w-t1,2)/2;        }         printf("Case #%d: ",kcase);        cout<<(S-p)/S<<endl;    }    return 0;}

UVA 11427 Expect the Expected

题意：每天晚上你会玩牌。第一把赢了就去睡觉，输了继续。你每一把赢的概率是p，当你赢的比例严格大于p时，才会去睡觉。但是你一个晚上最多玩n把游戏，如果哪天获胜比例一直没有大于p，你以后再也不玩了。求玩游戏天数的期望。

假设单独一天获胜概率为Q
则Ex=Q+2(1−Q)Q+3(1−Q)2Q+...
高中数学作差法可得Ex=1/Q

计算Q可以dp，dpi,j表示前i天赢j局时，最后‘垂头丧气′的概率
dp[i,j]=dp[i−1,j](1−p)+dp[i−1,j−1]p(j/i≤p)

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} const int maxn = 100 + 10;int n;double dp[maxn][maxn];int main(){//  freopen("uva11427.in","r",stdin);//  freopen(".out","w",stdout);    int T=read();    For(kcase,T) {        int a,b;        scanf("%d/%d %d",&a,&b,&n);        double p=(double)a/b;        MEM(dp) dp[0][0]=1;        For(i,n)            Rep(j,i)             {                if ((double)j/i>p) break;                dp[i][j]=dp[i-1][j]*(1-p)+( j ? dp[i-1][j-1]*p : 0 );             }         double ans=0;        Rep(j,n) ans+=dp[n][j];        printf("Case #%d: %d\n",kcase,(int)(1/ans));     }    return 0;}

UVA 11762 Race to 1

其实直接用马尔可夫过程就好

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} const int MAXN = (1000000+10);int primes[MAXN],size=0;bool b[MAXN]={0};void make_prime() {    int n=MAXN-10;    Fork(i,2,n) {        if (!b[i]) primes[++size]=i;        For(j,size) {            if (i*primes[j]>n) break;            b[i*primes[j]]=1;            if (i%primes[j]==0) break;        }    }}double dp[MAXN]={0};bool vis[MAXN]={0,1};double calc(int x) {    if (vis[x]) return dp[x];    dp[x]=1;     int siz=0,t=0;    double c=0;    for(int i=1;primes[i]<=x&&i<=size;++i) {        ++siz;        if (x%primes[i]==0) ++t,c+=calc(x/primes[i]);    }    return dp[x] = (c+siz)/t;   }int main(){//  freopen("uva11762.in","r",stdin);//  freopen(".out","w",stdout);    make_prime();    int T=read();    For(kcase,T) {        double ans=calc(read());        printf("Case %d: %.8lf\n",kcase,ans);    }    return 0;}

UVA 10491 So you want to be a 2n-aire?

很容易搞错的是，每次拿到题目后’我’马上知道答对概率，要根据这个概率决策的。

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define eps (1e-9)#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} int n;double t,p,dp[50];void work() {    dp[n]=(ll)1<<(ll)n;    if (fabs(1-t)<eps) {        cout<<dp[n]<<endl;         return;    }    RepD(i,n-1) {        double p2=((ll)(1<<(ll)i));        double f=p2 / dp[i+1];        if (f<=t) dp[i]=p*dp[i+1];        else dp[i] = ( (f-t)*p2 + (1-f)*(f+1) /2*dp[i+1] ) / (1-t);     }    cout<<dp[0]<<endl;}int main(){//  freopen("uva10900.in","r",stdin);//  freopen(".out","w",stdout);    cout<<setiosflags(ios::fixed)<<setprecision(3);    while(cin>>n>>t && n) {         p=(1+t)/2;        work();    }    return 0;}

UVA 11346 Probability

积分求面积，SB题

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} int main(){//  freopen("uva11346.in","r",stdin);    int T=read();    double a,b,S;    while(cin>>a>>b>>S) {        if (a*b<=S) {            puts("0.000000%");        } else if (S<1e-7) {            puts("100.000000%");        }else {            printf("%.6lf%\n",100-(1+log(a)-log(S/b))*S/a/b*100);        }    }       return 0;}

UVA 11637 Garbage Remembering Exam

期望具有线性性质，原题可以拆成每个单词无效无效概率的和，取反计算有效概率，线状排列中与第i个有效的概率为与在线排列中其所有距离不超过k的在都在环中超过k个的概率，暴力统计

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} const int MAXN = 100000 + 10;int k,n;double p[MAXN];double solve(int n,int k) {//  {2*k+1..n-1-2k}    MEM(p)     int x=2*k+1;    p[n]=1;    For(i,2*k)    {        p[n]=p[n]*(n-2*k+i)/(n-2*k-1+i);    }    ForkD(X,x,n) {        p[X-1]=p[X]*(X-2*k)/X;     }       double ans=0;    For(i,n) {        int len=n-(min(i+k,n)-max(i-k,1)+1);        ans+=1-p[len];    }     return ans;}int main(){//  freopen("uva11637.in","r",stdin);//  freopen(".out","w",stdout);    int kcase=1;    while (cin>>n>>k && n && k) {        double ans=0;        if (n==1) ans=0;        else if (n-1-2*k>0) {            ans=solve(n,k);        } else {            ans=n;        }           printf("Case %d: %.4lf\n",kcase++,ans);    }    return 0;}

UVA 11605 Lights inside a 3d Grid

用线性期望转成所有灯开的概率，由于这题是用xor和
根据
{(1−p+p)n=C0npn+C1n−1(1−p)pn−1+...+Cnn(1−p)n(1−p−p)n=(−1)nC0npn+(−1)n−1C1n−1(1−p)pn−1+...+Cnn(1−p)n
可以求出C1n(1−p)n−1p+C3n(1−p)n−3p3+...+Ckn(1−p)n−kpk∣k是奇数

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} int M,N,P,K;void calc(int i,int j,int k) {}int main(){//  freopen("uva11605.in","r",stdin);//  freopen(".out","w",stdout);    int T=read();    For(kcase,T) {        double ans=0;        cin>>N>>M>>P>>K;        For(i,N)            For(j,M)                For(k,P) {                    double p=(2.0*i*(N-i+1)-1)*(2*j*(M-j+1)-1)*(2*k*(P-k+1)-1)/(double)N/N/M/M/P/P;                    ans+=1-pow(1.0-2*p,K);                }        printf("Case %d: %.10lf\n",kcase,ans/2);    }    return 0;}

UVA 10491 Cows and Cars

注意所有门是一次打开的

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} int main(){//  freopen("uva11181.in","r",stdin);//  freopen(".out","w",stdout);    double a,b,c;    while(cin>>a>>b>>c) {        cout<<setiosflags(ios::fixed)<<setprecision(5);        cout<< (a*b+b*(b-1))/(a+b)/(a+b-c-1)<<endl;    }    return 0;}

UVA 1413 Game

卡精度问题，首先可以用区间dp算出一个某个区间的某侧人拿球时传左，右的概率，然后考虑第k个人拿球时，出现某个区间某侧人拿球的期望，只要n-1个人拿球了，后面无论如何都是n赢

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} const int MAXN = 50 +10;int n,k;double p[MAXN]; const int L = 0, R = 1;ld f[MAXN][MAXN][2][2]; // [left,Right,Ball,Excepted Direction]ld dp[MAXN][MAXN][2]; // P[Left,Right,Excepted Direction]int main(){//  freopen("uva1413.in","r",stdin);//  freopen(".out","w",stdout);    while(scanf("%d%d",&n,&k)==2) {        MEM(p) MEM(f) MEM(dp)        For(i,n-1) scanf("%lf",&p[i]);        For(i,n-1) {            f[i][i][L][L]=f[i][i][R][L]= 1 - p[i];            f[i][i][L][R]=f[i][i][R][R]= p[i];        }        // 用等比数列算出的公式 ：在区间[I,J]，球在k端时，从l端出去的概率          For(d,n) {            For(i,n-d-1) {                int j=i+d;                f[i][j][L][L]= (1-p[i]) / ( 1 - p[i] + p[i]*f[i+1][j][L][R] );                f[i][j][L][R]=  p[i]*f[i+1][j][L][R] / ( 1 - p[i] + p[i]*f[i+1][j][L][R] ); //如果直接用上面的概率减会卡精度                f[i][j][R][R]= (p[j]) / ( p[j] + (1 - p[j]) * f[i][j-1][R][L] );                f[i][j][R][L]=  ( (1 - p[j]) * f[i][j-1][R][L]) / ( p[j] + (1 - p[j]) * f[i][j-1][R][L] );            }        }        // 在区间[I,J]，球在k端的概率         dp[k][k][R]=1;        For(d,n-1) {            For(i,n-d-1) {                int j=i+d;                dp[i][j][L] = dp[i+1][j][L] * f[i+1][j][L][L] + dp[i+1][j][R] * f[i+1][j][R][L];                dp[i][j][R] = dp[i][j-1][L] * f[i][j-1][L][R] + dp[i][j-1][R] * f[i][j-1][R][R];            }        }        double ans = dp[1][n-1][L] + dp[1][n-1][R];        cout<<setiosflags(ios::fixed)<<setprecision(10);        cout<<ans<<endl;    }    return 0;}

UVA 1498 Activation

排队，注意各种特判，fi,j表示队列有i人排第j时“倒霉”的机率，先解方程求出fi,i递推。

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} const int MAXN = (2000+10); int N,M,K;double p1,p2,p3,p4;double f[MAXN][MAXN];  void prework() {    For(i,N) {        double k=p2/(1-p1);        f[i][i]=0;        double k2=pow(k,i-1),b2=0;        Fork(j,2,i) b2=b2*k+(p3*f[i-1][j-1]+p4 * (bool)(j<=K) )/ (1-p1);        f[i][i]= (k2 * p4 / (1-p1) + b2 ) / ( 1 - k2 * p2 / (1 - p1) ) ;        f[i][1]=( p2 * f[i][i] + p4 ) / ( 1 - p1 );        Fork(j,2,i-1) {            f[i][j] = (p2 * f[i][j-1] + p3 * f[i-1][j-1] + p4 * (j<=K) ) / (1-p1) ;        }    }}int main(){//  freopen("uva1498.in","r",stdin);//  freopen(".out","w",stdout);    while(cin>>N>>M>>K) {        cin>>p1>>p2>>p3>>p4;        if (p3 == 0 && p4 == 0 || p1 == 1 || p3 == 1 || p2 == 1 ) {            puts("0.00000"); continue;        }         MEM(f)        prework();        cout<<setiosflags(ios::fixed)<<setprecision(5);        cout<<f[N][M]<<endl;     }    return 0;}