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; }