C. Little Pony and Expected Maximum
2014-08-17 23:46
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这道题目的大意是,一个骰子有6个面分别写着1,2,3,4,5,6出现的概率相同,然后你可以扔n次,要求的是平均下来,可能的最大的数目。什么意识?比如一个两面的骰子,你可以投两次,下面是投的两次的几种可能:
1. You can get 1 in the first toss, and 2 in the second.Maximum equals to 2.
2. You can get 1 in the first toss, and 1 in the second.Maximum equals to 1.
3. You can get 2 in the first toss, and 1 in the second.Maximum equals to 2.
4. You can get 2 in the first toss, and 2 in the second.Maximum equals to 2
然后呢你就可以有:The probability of each outcome is 0.25, that is expectationequals to:
(2+1+2+2)*0.25(也就是1/4每一种情况是十分之一概率)。
上面标注的是组合数学,数学,的确是数学,但是运用组合数学的知识我并没有解出来,后来还是找规律写出来的,对于一个特定的面数的骰子,每一种出现的可能次数是固定的,比如说6面,一共36种情况那么:1最大的是1种,2最大的是上面三种,三最大的是5种(9-1-3)为什么?因为3最大就不需要考虑4,5,6了,所以就可以是3个面的了,一共是9次,那么4也是一样的,也就是他的这个次数和面数以及次数有一个平行走向的关系。
在这一题目中如果有幂次的运用如果你用循环就会超时,所以还是用pow函数直接AC
1. You can get 1 in the first toss, and 2 in the second.Maximum equals to 2.
2. You can get 1 in the first toss, and 1 in the second.Maximum equals to 1.
3. You can get 2 in the first toss, and 1 in the second.Maximum equals to 2.
4. You can get 2 in the first toss, and 2 in the second.Maximum equals to 2
然后呢你就可以有:The probability of each outcome is 0.25, that is expectationequals to:
(2+1+2+2)*0.25(也就是1/4每一种情况是十分之一概率)。
上面标注的是组合数学,数学,的确是数学,但是运用组合数学的知识我并没有解出来,后来还是找规律写出来的,对于一个特定的面数的骰子,每一种出现的可能次数是固定的,比如说6面,一共36种情况那么:1最大的是1种,2最大的是上面三种,三最大的是5种(9-1-3)为什么?因为3最大就不需要考虑4,5,6了,所以就可以是3个面的了,一共是9次,那么4也是一样的,也就是他的这个次数和面数以及次数有一个平行走向的关系。
在这一题目中如果有幂次的运用如果你用循环就会超时,所以还是用pow函数直接AC
#include<iostream> using namespace std; #include<cstring> #include<stdio.h> #include<cstdlib> #include<cmath> int m,n; /* double mi(int M,int n) { double k1=1.0,k2=1.0; for(int i=1;i<=n;i++) k1*=M*1.0/m; for(int i=1;i<=n;i++) k2*=(M-1)*1.0/m; return k1-k2; } */ int main() { int i,j,k; double sum; while(cin>>m>>n) { sum=0; for(i=1;i<=m;i++) { // cout<<mi(i,n)*i<<endl; sum+=(pow(i*1.0/m,n)-pow((i-1)*1.0/m,n))*i; } printf("%.12lf\n",sum); } return 0; }
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