标签：stream into tput printf blog any eal ant put
Alice and Bob are playing a simple game. They line up a row of nn identical coins, all with the heads facing down onto the table and the tails upward.
For exactly mm times they select any kk of the coins and toss them into the air, replacing each of them either heads-up or heads-down with the same possibility. Their purpose is to gain as many coins heads-up as they can.
The input has several test cases and the first line contains the integer t (1 \le t \le 1000)t(1≤t≤1000) which is the total number of cases.
For each case, a line contains three space-separated integers nn, m (1 \le n, m \le 100)m(1≤n,m≤100)and k (1 \le k \le n)k(1≤k≤n).
For each test case, output the expected number of coins heads-up which you could have at the end under the optimal strategy, as a real number with the precision of 33 digits.
6 2 1 1 2 3 1 5 4 3 6 2 3 6 100 1 6 100 2
0.500 1.250 3.479 3.000 5.500 5.000
题意：有n枚朝下的硬币，我们可以投掷这些硬币m次，每次投掷 t 枚硬币，问最后朝上硬币的期望
分析：最优的策略一定是：当有至少 k 枚硬币面朝下时，则选 k 枚面朝下的硬币去抛掷（任意k 枚都可以）；如果不足 k 枚面朝下，则在选择所有面朝下的硬币的基础上再额外选择若干面朝上的硬币。
于是有动态规划，记 dp[i][j]表示抛掷了 i 次后，有 j 枚硬币面朝上的概率。他们应该满足dp[i][0]+dp[i][1]+...+dp[i][n]=1。转移时，考虑从当前状态(i,j)出发，抛掷的 k 枚硬币的所有可能结果：分别有 0~k 枚面朝上。其中 k 枚硬币抛掷后有 l 枚面朝上的概率为 C(k,l)/2k。时间复杂度 O(nmk)。
参考博客：https://blog.csdn.net/mitsuha_/article/details/79307065
AC代码：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 205;
const ll mod = 1e9 + 7;
double dp[maxn][maxn], p[maxn], c[maxn][maxn];
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
c[0][0] = 1;
for( ll i = 1; i <= 100; i ++ ) {
c[i][0] = 1;
for( ll j = 1; j <= i; j ++ ) {
c[i][j] = c[i-1][j-1] + c[i-1][j]; //打表前一百个的组合数
}
}
p[0] = 1;
for( ll i = 1; i <= 100; i ++ ) {
p[i] = p[i-1]/2; //几枚硬币朝上的概率
}
ll T;
cin >> T;
while( T -- ) {
ll n, m, t;
cin >> n >> m >> t;
memset( dp, 0, sizeof(dp) );
dp[0][0] = 1; //记录投掷i次有j枚硬币朝上的概率
for( ll i = 0; i < m; i ++ ) {
for( ll j = 0; j <= n; j ++ ) {
if( dp[i][j] == 0 ) {
continue;
}
for( ll k = 0; k <= t; k ++ ) {
if( n-j >= t ) { //还有硬币没有朝上的情况
dp[i+1][j+k] += dp[i][j]*c[t][k]*p[t];
} else { //已经有n枚硬币朝上了还得投掷的情况，这时会使n枚变少或者不变
dp[i+1][n-t+k] += dp[i][j]*c[t][k]*p[t]; //n-t代表会改变t枚硬币的情况，k代表改变的情况朝上的情况
}
}
}
}
double ans = 0;
for( ll i = 1; i <= n; i ++ ) {
ans += dp[m][i]*i; //计算期望
}
printf("%.3lf\n",ans);
}
return 0;
}
ACM-ICPC 2017 Asia Urumqi：A. Coins（DP） 组合数学
标签：stream into tput printf blog any eal ant put
原文地址：https://www.cnblogs.com/l609929321/p/9387663.html
