hdoj f(n)2582(GCDは時計を打って&法則を探します)良い問題を探します
2012 ワード
f(n)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 378 Accepted Submission(s): 230
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
Sample Output
表を打ってGCDの法則を探すと、GCD(n)は、nが1つの質量因子しかない場合、GCD(n)は1ではなく、このときGCDはnのこの唯一の質量因子に等しいことがわかります.
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 378 Accepted Submission(s): 230
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
3
26983
Sample Output
3
37556486
表を打ってGCDの法則を探すと、GCD(n)は、nが1つの質量因子しかない場合、GCD(n)は1ではなく、このときGCDはnのこの唯一の質量因子に等しいことがわかります.
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll __int64
#define N 1000010
using namespace std;
ll p[N];
bool flag[N];
ll sum[N];
void init()
{
__int64 i,j,num=0;
for(i=2;i<=N;i++)
{
if(!flag[i])
{
p[num++]=i;
for(j=i*i;j<=N;j+=i)
flag[j]=true;
}
}
}
ll solve(ll n)
{
ll i,k,ans=0;
for(i=0;p[i]*p[i]<=n;i++)
{
if(n%p[i]==0)
{
n/=p[i];
while(n%p[i]==0)
n/=p[i];
k=p[i];
ans++;
}
if(ans>=2)
return 1;
}
if(n>1)
{
ans++;
k=n;
}
if(ans>=2)
return 1;
else
return k;
}
int main()
{
ll n,i;
init();
for(i=3;i<=N;i++)
{
if(flag[i])
sum[i]=sum[i-1]+solve(i);
else
sum[i]=sum[i-1]+i;
}
while(scanf("%d",&n)!=EOF)
printf("%I64d
",sum[n]);
return 0;
}