题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1012
解法一:最直观的方法,就是递归计算来求各项之和
#include <iostream>
#include <iomanip>
using namespace std;
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double curItem(int n)
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{//当前子项
if (n==0)
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{
return 1;
}
else
return curItem(n-1)/n;
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}
double sum(int n)
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{
if(n==0)
return 1;
else
return sum(n-1)+curItem(n);
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}
void caculateE(int n)
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{//输出指定n下e的值
double tmp1,tmp2=1.0f,result=0.0f;
int i;
result = sum(n);
if(n==0||n==1)
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{
cout<<n<<" "<<static_cast<int>(result)<<endl;
}
else if(n==2)
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{
cout<<setiosflags(ios::fixed)<<setprecision(1);
cout<<n<<" "<<result<<endl;
}
else
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{
cout<<setiosflags(ios::fixed)<<setprecision(9);
cout<<n<<" "<<result<<endl;
}
}
int main(int argc, char *argv[])
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{
int n,i;
cout<<"n e"<<endl;
cout<<"- -----------"<<endl;
for(i=0;i<=9;++i)
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{
caculateE(i);
}
cin>>i;
return 0;
}
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解法二:题中给出的计算e的式子是由e^x的泰勒级数展开而得,在计算之前可以使用个技巧,就是把它们叠乘起来改写成:
e=(1+(1+1/2(1+1/3(1+1/4(1+…1/(n-1)(1+1/n))))),从最里面的括号往外算,共做n次除法和加法得一段结果,运算效率也是O(N*M),但是由于收敛速度快些,所以N项节省一些,
#include <iostream>
#include <iomanip>
using namespace std;
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double doCaculate(int n)
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{//实际的计算
double tmp1,tmp2=1.0f,result=0.0f;
int i;
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for(i=n;i>=1;--i)
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{
tmp1 = static_cast<double>(1)/static_cast<double>(i);
tmp2 = 1.0f+tmp1*tmp2;
}
return tmp2;
}
void caculateE(int n)
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{//输出指定n下e的值,主要是输出格式的处理
double result = 0.0f;
result = doCaculate(n);
if(n==0||n==1)
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{
cout<<n<<" "<<static_cast<int>(result)<<endl;
}
else if(n==2)
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{
cout<<setiosflags(ios::fixed)<<setprecision(1);
cout<<n<<" "<<result<<endl;
}
else
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{
cout<<setiosflags(ios::fixed)<<setprecision(9);
cout<<n<<" "<<result<<endl;
}
}
int main(int argc, char *argv[])
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{
int n,i;
cout<<"n e"<<endl;
cout<<"- -----------"<<endl;
for(i=0;i<=9;++i)
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{
caculateE(i);
}
return 0;
}
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