hdu 4135 Co-prime 容斥原理

Co-prime

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

[align=left]Problem Description[/align]
Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.

[align=left]Input[/align]
The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 1015) and (1 <=N <= 109).

[align=left]Output[/align]
For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.

[align=left]Sample Input[/align]

2
1 10 2
3 15 5

[align=left]Sample Output[/align]

Case #1: 5
Case #2: 10

Hint

In the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.

[align=left]Source[/align]
The Third Lebanese Collegiate Programming Contest
题意：给你三个数，求[A,B]区间内与N互质的数的个数；
思路：即求[1,B]-[1,A-1]内与N互质的数
　　　因为n为1e9,所以根号打表，将N唯一分解，
　　　容斥得到与N不互质的数的个数；
　　　ans=B-[1,B]-(A-1-[1,A-1]);

#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 ;
}
ll prime[100010];
ll vis[100010];
ll a[110];
ll ji,cnt;
ll ans,x,y,z;
ll gcd(ll x,ll y)
{
return y==0?x:gcd(y,x%y);
}
void Prime(ll n)
{
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;
}
}
}
void dfs(ll lcm,ll pos,ll step,ll x,ll &ans)
{
if(lcm>x)
return;
if(pos==ji)
{
if(step==0)
return;
if(step&1)
ans+=(x/lcm);
else
ans-=(x/lcm);
return;
}
dfs(lcm,pos+1,step,x,ans);
dfs(lcm/gcd(a[pos],lcm)*a[pos],pos+1,step+1,x,ans);
}
int main()
{
ll t,i;
int cs=1;
Prime(100000);
scanf("%I64d",&t);
while(t--)
{
ji=0;
scanf("%I64d%I64d%I64d",&x,&y,&z);
for(i=0;i<cnt;i++)
if(z%prime[i]==0)
{
a[ji++]=prime[i];
while(z%prime[i]==0)
z/=prime[i];
}
if(z>1)
a[ji++]=z;
ll gg=0;
ans=0;
dfs(1,0,0,x-1,gg);
dfs(1,0,0,y,ans);
printf("Case #%d: ",cs++);
printf("%I64d\n",(y-ans)-(x-1-gg));
}
return 0;
}

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