求1+2+3+4+...+n的和
n >= 1 && n <= 1,000,000,000
#include<iostream>
using namespace std;
int main()
{
long long int s,count = 0;
cin>>s;
for(;s >= 1;s--)
{
count += s;
}
cout<<count<<endl;
system("pause");
cin.get();
return 0;
}
等差数列求和公式:S=[项数 *(首项 + 尾项)] / 2
#include<iostream>
using namespace std;
int main()
{
long long int s;
cin>>s;
cout<<s*(1+ s)/2<<endl;
system("pause");
return 0;
}
#include<iostream>
using namespace std;
long long int fun(long long int s)
{
if(s > 0)
{
return s + fun(s - 1);
}
else
return 0;
}
int main(int argc, char const *argv[])
{
long long int s,count;
cin>>s;
count = fun(s);
cout<<count<<endl;
return 0;
}
原文:https://www.cnblogs.com/l574/p/14984356.html