题意：要求输入一篇N个字符的文章，对所有非负整数i：每到第i+0.1秒时可以输入一个文章字符，每到第i+0.9秒时有P的概率崩溃（回到开头或者上一个存盘点）
每到第i秒有一次机会可以选择按下X个键存盘，或者不存，打印完整篇文章之后必须存盘一次才算完成输入多组N,P,X选择最佳策略使得输入完整篇文章时候按键的期望最小，
输出此期望
析：dp[i]表示打完前 i 个字符，概论是多少，dp[i] = dp[i-1] + p(1+dp[i]) + 1-p。然后解得dp[i] = (dp[i-1]+1) / (1-p)。
最后再枚举多少次保存。均匀分布是最优的。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
double dp[maxn]; int main(){
int T; cin >> T;
for(int kase = 1; kase <= T; ++kase){
double p;
scanf("%d %lf %d", &n, &p, &m);
dp[0] = 0;
for(int i = 1; i <= n; ++i) dp[i] = (dp[i-1]+1) / (1-p);
double ans = inf;
for(int i = 1; i <= n; ++i){
int a = n / i;
int b = n % i;
ans = min(ans, dp[a+1] * b + dp[a] * (i-b) + (double)m*i);
}
printf("Case #%d: %f\n", kase, ans);
}
return 0;
}