Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 457 Accepted Submission(s): 279
f(n)= Gcd(3)+Gcd(4)+…+Gcd(i)+…+Gcd(n).
Gcd(n)=gcd(C[n][1],C[n][2],……,C[n][n-1])
C[n][k] means the number of way to choose k things from n some things.
gcd(a,b) means the greatest common divisor of a and b.
The output consists of one line with one integer f(n).
n为素数:本身
n有多个素因子:1
n只有一个素因子:该因子
920MS才跑过去。。没优化了。。记得用_int64保存结果。
#include <stdio.h> #include <string.h> #include <algorithm> #include <iostream> using namespace std; typedef long long LL; const int N = 1000001; bool p[N]; LL g[N]; LL f[N]; void init(){ memset(g,0,sizeof(g)); for(int i=2;i<N;i++){ if(!p[i]){ g[i]=(LL)i; for(LL j = (LL)i*i;j<N;j+=i){ p[j] = true; } } } f[2]=0; for(int i=3;i<N;i++){ int cnt = 0; if(g[i]==0){ int temp=-1; int n = i; for(int j=2;j*j<=n;j++){ if(n%j==0){ temp=(LL)j; cnt++; while(n%j==0){ n/=j; } } } if(n>1) {temp=(LL)n,cnt++;} if(cnt!=1) g[i]=1; if(cnt==1) g[i]=(LL)temp; } f[i]=f[i-1]+g[i]; } } int main() { init(); int n; while(~scanf("%d",&n)){ printf("%lld\n",f[n]); } return 0; }