So you want to be a 2n-aire? UVA - 10900
2017-11-10 14:57
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当遇到了当前的这道题目之后有两种选择。一种是直接放弃,这时候会得到对应的钱,另外一种是选择继续做答,如果这时候得到的钱是答对当前的题目的基础上,从后一道题目继续作答所得到的钱的最大期望值。对应于这两种情况,都有对应的式子,根据大小进行判断得到一个临界的概率值,同时这个概率值不能够小于t,然后就可以计算出当前得到的钱的最大的期望值,从而建立递推关系式得到最终的结果,具体实现见如下代码:
#include<iostream>
#include<vector>
#include<string>
#include<set>
#include<stack>
#include<queue>
#include<map>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<deque>
#include<functional>
using namespace std;
double d[35];
int main(){
int n;
double t;
while (cin >> n >> t){
if (n == 0 && t == 0) break;
d
= 1 << n;
for (int i = n - 1; i >= 0; i--){
double p0 = (1 << i) / d[i + 1];
if (p0 < t) p0 = t;
double p1 = (p0 - t) / (1 - t);
d[i] = (1 << i)*p1 + (1 + p0) / 2 * d[i + 1] * (1 - p1);
}
cout << fixed << setprecision(3) << d[0] << endl;
}
return 0;
}
#include<iostream>
#include<vector>
#include<string>
#include<set>
#include<stack>
#include<queue>
#include<map>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<deque>
#include<functional>
using namespace std;
double d[35];
int main(){
int n;
double t;
while (cin >> n >> t){
if (n == 0 && t == 0) break;
d
= 1 << n;
for (int i = n - 1; i >= 0; i--){
double p0 = (1 << i) / d[i + 1];
if (p0 < t) p0 = t;
double p1 = (p0 - t) / (1 - t);
d[i] = (1 << i)*p1 + (1 + p0) / 2 * d[i + 1] * (1 - p1);
}
cout << fixed << setprecision(3) << d[0] << endl;
}
return 0;
}
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