hdu 1178 Heritage from father

求Sn=(1*2+2*3+....+n*(n+1))/2;

因为n*(n+1)=n^2+n，又1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6  1+2+3+...+n=n(n+1)/2;

所以S=[n(n+1)(2n+1)/6+n(n+1)/2]/2=n(n+1)(n+2)/6;

//n*(n+1)*(n+2)/6=m*10^a//log(n*(n+1)*(n+2)/6)=logm+a//a=[log(n*(n+1)*(n+2)/6)]取整//m=10^(log(n*(n+1)*(n+2)/6)-a)#include<iostream>#include<cmath> #include<iomanip>using namespace std;int main(){double n,a,m,b;while(cin>>n&&n){b=log10(n*1.0/6)+log10((n+1)*1.0)+log10((n+2)*1.0);a=b-int(b);m=pow(double(10),a);printf("%0.2lfE%d/n",m,int(b));} return 0;}