Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) D. Jon and Orbs 概率DP

D. Jon and Orbs

链接：

http://codeforces.com/contest/768/problem/D

题解：

设dp[i][j]为第ｉ天时已经产生ｊ种龙晶，转移方程为

dp[i][j] += dp[i - 1][j] * j / k;
dp[i][j + 1] += dp[i - 1][j] * (k - j) / k;

代码：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1000000007;
const double eps = 1e-7;
// head

double dp[10005][1005];
int main()
{
    int k, q;
    scanf("%d%d", &k, &q);
    dp[0][0] = 1.0;
    rep(i, 1, 10000) rep(j, 0, k) { 
            dp[i][j] += dp[i - 1][j] * j / k;
            dp[i][j + 1] += dp[i - 1][j] * (k - j) / k;
        }
    rep(i, 1, q) {
        int p;
        scanf("%d", &p);
        rep(j, 0, 10000)
            if (dp[j][k] > (p - eps) / 2000){
                printf("%d
", j);
                break;
            }
    }
    return 0;
}