hdu 2582(数论相关定理+素数筛选+整数分解)

f(n)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 457 Accepted Submission(s): 279


[align=left]Problem Description[/align]
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
[1],C
[2],……,C
[n-1])
C
[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.

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

[align=left]Output[/align]
For each test case:
The output consists of one line with one integer f(n).

[align=left]Sample Input[/align]

3
26983

[align=left]Sample Output[/align]

3 37556486
这题要用到这个公式:
设

，那么G的值为：

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
;
LL g
;
LL f
;
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
);
}
return 0;
}