hdu 2588 gcd 欧拉函数

GCD

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)


[align=left]Problem Description[/align]
The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.

[align=left]Input[/align]
The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (2<=N<=1000000000, 1<=M<=N), representing a test case.

[align=left]Output[/align]
For each test case,output the answer on a single line.

[align=left]Sample Input[/align]

3
1 1
10 2
10000 72

[align=left]Sample Output[/align]

1
6
260

[align=left]Source[/align]
ECJTU 2009 Spring Contest
思路：跟poj 2480 差不多；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
int res = 0 , ch ;
while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
{
if( ch == EOF ) return 1 << 30 ;
}
res = ch - '0' ;
while( ( ch = getchar() ) >= '0' && ch <= '9' )
res = res * 10 + ( ch - '0' ) ;
return res ;
}
#define MAXN 100001
ll prime[MAXN];//保存素数
ll vis[MAXN],ji;//初始化
ll Prime(ll n)
{
ll cnt=0;
//memset(vis,0,sizeof(vis));
for(ll i=2;i<=n;i++)
{
if(!vis[i])
prime[cnt++]=i;
for(ll j=0;j<cnt&&i*prime[j]<n;j++)
{
vis[i*prime[j]]=1;
if(i%prime[j]==0)//关键
break;
}
}
return cnt;
}
ll phi(ll n)
{
ll i,rea=n;
for(i=0;i<ji;i++)
{
if(prime[i]*prime[i]>n)break;
if(n%prime[i]==0)
{
rea=rea-rea/prime[i];
while(n%prime[i]==0)  n/=prime[i];
}
}
if(n>1)
rea=rea-rea/n;
return rea;
}
int main()
{
ll x,y,z,i,t;
ji=Prime(52010);
int T;
scanf("%d",&T);
while(T--)
{
scanf("%I64d%I64d",&x,&y);
ll ans=0;
for(i=1;i*i<=x;i++)
{
if(x%i==0)
{
ll gg=i;
ll hh=x/i;
if(gg>=y)
ans+=phi(hh);
if(gg!=hh&&hh>=y)
ans+=phi(gg);
}
}
printf("%I64d\n",ans);
}
return 0;
}

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