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.

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1010
 #define maxv 1010
 #define mod 1000000000
 using namespace std;

 double solve(int n,int m)
 {
     double ans=;
     for(int i=n;i>m;i--)
         ans*=i;
     for(int i=;i<=n-m;i++)
         ans/=i;
     return ans;
 }

 int main()
 {
    // Read();
     int t,n,m;
     double x,ans;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d%lf",&n,&m,&x);
         ans=;
         for(int i=m;i<=n;i++)
         {
             ans+=solve(n,i)*pow(x,i)*pow((-x),(n-i));
             //printf("%.4lf %.4lf\n",pow(x,i),pow(1-x,n-i));
         }
         printf("%.4lf\n",ans);
     }
     return ;
 }