HDU 2582 规律题

 

f(n)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 165    Accepted Submission(s): 92

Problem Description

This time I need you to calculate the f(n) . (3<=n<=1000000)

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.

 

Input

There are several test case. For each test case:One integer n(3<=n<=1000000). The end of the in put file is EOF.

 

Output

For each test case:
The output consists of one line with one integer f(n).

 

Sample Input

326983

 

Sample Output

337556486

 

Source

ECJTU 2009 Spring Contest

 

Recommend

lcy

 

直接推，真的很难推出什么名堂。。

但是通过打表找GCD的规律的话，你会发现如下规律：
GCD(n)，当n只有一个质因子的时候GCD(n)不会为1。

此时GCD等于n的这个唯一的质因子。

 

所以预处理的时候，只需要判断i是不是可以写成p^k就行了

顺便说一句，必须要用__int64才能存的下。

 

我的代码：

#include<stdio.h>__int64 prime[100000];bool flag[1000001];__int64 sum[1000001];void init(){__int64 i,j,num=0;for(i=2;i<=1000000;i++){if(!flag[i]){prime[num++]=i;for(j=i*i;j<=1000000;j=j+i)flag[j]=true;}}}__int64 solve(__int64 n){__int64 i,k,ret=0;for(i=0;prime[i]*prime[i]<=n;i++){if(n%prime[i]==0){n=n/prime[i];while(n%prime[i]==0)n=n/prime[i];k=prime[i];ret=ret+1;}if(ret>=2)return 1;}if(n>1){ret=ret+1;k=n;}if(ret>=2)return 1;elsereturn k;}int main(){__int64 n,i;init();for(i=3;i<=1000000;i++){if(flag[i])sum[i]=sum[i-1]+solve(i);elsesum[i]=sum[i-1]+i;}while(scanf("%d",&n)!=EOF)printf("%I64d\n",sum[n]);return 0;}