【[Offer收割]编程练习赛 14 B】投掷硬币

【题目链接】:http://hihocoder.com/problemset/problem/1506

【题意】

中文题

【题解】

这种题是概率DP….
设f[i][j]表示i个硬币里面有j个正面朝上的概率;
则第i个有两种可能;
证明朝上或反面朝上;
f[i][j]=f[i-1][j]*(1-p[i])+f[i-1][j-1]*p[i];

【Number Of WA】

0

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 1100;

int n,m;
double p[N],f[N][N];

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    ios::sync_with_stdio(false);
    cin >> n >> m;
    rep1(i,1,n)
        cin >> p[i];
    double allzheng=1,allfan=1;
    rep1(i,1,n)
    {
        allzheng*=p[i];
        allfan*=(1-p[i]);
        f[i][i] = allzheng,f[i][0]=allfan;
    }
    rep1(i,2,n)
        rep1(j,1,min(i,m))
                f[i][j] = f[i-1][j]*(1-p[i])+f[i-1][j-1]*p[i];
    cout << fixed << setprecision(9) << f[n][m] << endl;
    //printf("
%.2lf sec 
", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}