UVA 10491-Cows and Cars
2012-10-25 13:18
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水题,可以这样想,一开始有m个cow,n个car,然后主持人打开k个门,由于主持人只会打开是cow的门,并且在他打开门之后仍然可以选择是否改变自己的选择;思路如下:
假如一开始选择的是cow ,则概率:cow/(cow+car);主持人打开门之后如果改的话改成car的概率:car/(car+cow-show-1);则选到car的概率为cow/(cow+car)*car/(car+cow-show-1);
假如一开始选择的是car,则概率:car/(cow+car);主持人打开门之后如果改的话改成car的概率:(car-1)/(car+cow-show-1);则选到car的概率为cow/(cow+car)*(car-1)/(car+cow-show-1);
总的概率为:cow/(car+cow)*(car/(car+cow-show-1))+car/(car+cow)*((car-1)/(car+cow-show-1));
由于要保留5位小数,所以用double存数据;
#include<cstdio>
int main()
{
double car,cow,show;
while(scanf(\"%lf%lf%lf\",&cow,&car,&show)!=EOF)
{
double ans=cow/(car+cow)*(car/(car+cow-show-1))+car/(car+cow)*((car-1)/(car+cow-show-1));
printf(\"%.5lf\\n\",ans);
}
return 0;
}
假如一开始选择的是cow ,则概率:cow/(cow+car);主持人打开门之后如果改的话改成car的概率:car/(car+cow-show-1);则选到car的概率为cow/(cow+car)*car/(car+cow-show-1);
假如一开始选择的是car,则概率:car/(cow+car);主持人打开门之后如果改的话改成car的概率:(car-1)/(car+cow-show-1);则选到car的概率为cow/(cow+car)*(car-1)/(car+cow-show-1);
总的概率为:cow/(car+cow)*(car/(car+cow-show-1))+car/(car+cow)*((car-1)/(car+cow-show-1));
由于要保留5位小数,所以用double存数据;
#include<cstdio>
int main()
{
double car,cow,show;
while(scanf(\"%lf%lf%lf\",&cow,&car,&show)!=EOF)
{
double ans=cow/(car+cow)*(car/(car+cow-show-1))+car/(car+cow)*((car-1)/(car+cow-show-1));
printf(\"%.5lf\\n\",ans);
}
return 0;
}
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