FZU Problem 2132 LQX的作业 (数学题)
http://acm.fzu.edu.cn/problem.php?pid=2132
N个数已经排成非递减顺序,那么每次可以取 前m->n个在x前面.
取前m个在x前面的概率是 C(n,m)*x^m*(1-x)^(n-m)
依次递推即可.
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <vector> 5 #include <cstring> 6 #include <string> 7 #include <algorithm> 8 #include <string> 9 #include <set> 10 #include <functional> 11 #include <numeric> 12 #include <sstream> 13 #include <stack> 14 #include <map> 15 #include <queue> 16 //#pragma comment(linker, "/STACK:102400000,102400000") 17 #define CL(arr, val) memset(arr, val, sizeof(arr)) 18 19 #define ll long long 20 #define inf 0x7f7f7f7f 21 #define lc l,m,rt<<1 22 #define rc m + 1,r,rt<<1|1 23 #define pi acos(-1.0) 24 25 #define L(x) (x) << 1 26 #define R(x) (x) << 1 | 1 27 #define MID(l, r) (l + r) >> 1 28 #define Min(x, y) (x) < (y) ? (x) : (y) 29 #define Max(x, y) (x) < (y) ? (y) : (x) 30 #define E(x) (1 << (x)) 31 #define iabs(x) (x) < 0 ? -(x) : (x) 32 #define OUT(x) printf("%I64d\n", x) 33 #define lowbit(x) (x)&(-x) 34 #define Read() freopen("a.txt", "r", stdin) 35 #define Write() freopen("b.txt", "w", stdout); 36 #define maxn 1010 37 #define maxv 1010 38 #define mod 1000000000 39 using namespace std; 40 41 double solve(int n,int m) 42 { 43 double ans=1; 44 for(int i=n;i>m;i--) 45 ans*=i; 46 for(int i=1;i<=n-m;i++) 47 ans/=i; 48 return ans; 49 } 50 51 int main() 52 { 53 // Read(); 54 int t,n,m; 55 double x,ans; 56 scanf("%d",&t); 57 while(t--) 58 { 59 scanf("%d%d%lf",&n,&m,&x); 60 ans=0; 61 for(int i=m;i<=n;i++) 62 { 63 ans+=solve(n,i)*pow(x,i)*pow((1-x),(n-i)); 64 //printf("%.4lf %.4lf\n",pow(x,i),pow(1-x,n-i)); 65 } 66 printf("%.4lf\n",ans); 67 } 68 return 0; 69 }
