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
", 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=1;
    for(int i=n;i>m;i--)
        ans*=i;
    for(int i=1;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=0;
        for(int i=m;i<=n;i++)
        {
            ans+=solve(n,i)*pow(x,i)*pow((1-x),(n-i));
            //printf("%.4lf %.4lf
",pow(x,i),pow(1-x,n-i));
        }
        printf("%.4lf
",ans);
    }
    return 0;
}