#15. 钻石

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<vector>
 5 using namespace std;
 6 int tt;
 7 int n,m;
 8 int v[35];
 9 double p[35];
10 double ans[35];
11 vector<pair<int,double> > sta[35];
12 int main(){
13 
14     scanf("%d%d",&n,&m);
15     for(int i=1,x;i<=n;i++){
16         scanf("%d%d",&v[i],&x);
17         p[i]=x/100.;
18     }
19     for(int i=0;i<=n;i++){
20         sta[i].clear();
21     }
22     int an=(n/2.5)+1;
23     int bn=n-an;
24     for(int st=0;st<1<<bn;st++){
25         double nowp=1;
26         int cnt=0,money=0;
27         for(int i=0;i<bn;i++){
28             if((st>>i)&1){
29                 money+=v[n-i];
30                 nowp*=p[n-i];
31             }else{
32                 cnt++;
33                 nowp*=(1-p[n-i]);
34             }
35         }
36         sta[cnt].push_back(make_pair(money,nowp));
37     }
38     for(int i=0;i<=n;i++){
39         sort(sta[i].begin(),sta[i].end());
40         for(int j=1;j<sta[i].size();j++){
41             sta[i][j].second+=sta[i][j-1].second;
42         }
43     }
44     for(int st=0;st<1<<an;st++){
45         double nowp=1;
46         int cnt=0,money=0;
47         for(int i=0;i<an;i++){
48             if((st>>i)&1){
49                 money+=v[i+1];
50                 nowp*=p[i+1];
51             }else{
52                 cnt++;
53                 nowp*=(1-p[i+1]);
54             }
55         }
56         for(int i=0;i<=bn;i++){
57             // now d =cnt+i
58             int L = m-money;
59             vector<pair<int,double> >::iterator it = lower_bound(sta[i].begin(),sta[i].end(),make_pair(L,-1.));
60             double tmp = sta[i].back().second;
61             if(it!= sta[i].begin()){
62                 it--;
63                 tmp-=it->second;
64             }
65             ans[cnt+i] += tmp*nowp;
66         }
67     }
68     for(int i=0;i<=n;i++){
69         printf("%.3f\n",ans[i]);
70     }
71      fclose(stdout);
72     return 0;
73 }

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<map>
 6 #include<algorithm>
 7 #define lli long long int 
 8 using namespace std;
 9 const int MAXN=88000;
10 inline void read(int &n)
11 {
12     char c=getchar();n=0;bool flag=0;
13     while(c<'0'||c>'9')    c=='-'?flag=1,c=getchar():c=getchar();
14     while(c>='0'&&c<='9')    n=n*10+c-48,c=getchar();flag==1?n=-n:n=n;
15 }
16 struct node
17 {
18     int p,v;
19 }box[MAXN];
20 double ans[MAXN];
21 int n,m;
22 int base;// 中间节点 
23 struct ANS1
24 {
25     int zuanshi;
26     int qian;
27     double gailv;
28 }ans1[MAXN];
29 int ans1tot=0;
30 struct ANS2
31 {
32     int zuanshi;
33     int qian;
34     double gailv;
35 }ans2[MAXN];
36 int ans2tot=0;
37 void dfs(int num,double gl,int money,int zs)
38 {
39     if(num==base)
40     {
41         ans1[++ans1tot].qian=money;
42         ans1[ans1tot].zuanshi=zs;
43         ans1[ans1tot].gailv=gl;
44         return ;
45     }
46     dfs(num+1,(double)gl*((double)box[num+1].p/100),money+box[num+1].v,zs);
47     dfs(num+1,(double)gl*(double)( 1 - ( (double) box[num+1].p/100) ),money,zs+1);
48 }
49 void dfs2(int num,double gl,int money,int zs)
50 {
51     if(num==n)
52     {
53         ans2[++ans2tot].qian=money;
54         ans2[ans2tot].zuanshi=zs;
55         ans2[ans2tot].gailv=gl;
56         return ;
57     }
58     dfs2(num+1,(double)gl*((double)box[num+1].p/100),money+box[num+1].v,zs);
59     dfs2(num+1,(double)gl*(double)( 1 - ( (double) box[num+1].p/100) ),money,zs+1);
60 }
61 int comp1(const ANS1 &a,const ANS1 &b)
62 {
63     return a.qian<b.qian;
64 }
65 int comp2(const ANS2 &a,const ANS2 &b)
66 {
67     return a.qian>b.qian;
68 }
69 int main()
70 {
71 
72     read(n);read(m);
73     base=n/2;
74     for(int i=1;i<=n;i++)
75     {
76         read(box[i].v);
77         read(box[i].p);
78     }
79     dfs(1,(double)box[1].p/100,box[1].v,0);// 钱 
80     dfs(1,(double)1-(double)box[1].p/100,0,1);// 钻石 
81     
82     dfs2(base+1,(double)box[base+1].p/100,box[base+1].v,0);// 钱 
83     dfs2(base+1,(double)1-(double)box[base+1].p/100,0,1);// 钻石 
84     
85     sort(ans1+1,ans1+ans1tot,comp1);
86     sort(ans2+1,ans2+ans2tot,comp2);
87     for(int i=1;i<=ans1tot;i++)
88     {
89         for(int j=1;j<=ans2tot;j++)
90         {
91             if(ans1[i].qian+ans2[j].qian>=m)
92                 ans[ans1[i].zuanshi+ans2[j].zuanshi]+=ans1[i].gailv*ans2[j].gailv;
93             else break;
94         }
95     }
96     for(int i=0;i<=n;i++)
97         printf("%.3lf\n",ans[i]);
98     return 0;
99 }