HDU 4135 Co-prime(素因数分解+反発原理)
Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6969 Accepted Submission(s): 2751
Problem Description
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.
Input
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).
Output
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.
Sample Input
2 1 10 2 3 15 5
Sample Output
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}.
AからBまでのnとの相互作用の数はどのくらいあるか.
1−Bにおけるnとの相互作用の個数と、1−(A−1)におけるnとの相互作用の個数を反発により求め、その後+1(1はいずれの数とも相互作用し、2回減算したので+1が必要)を減算すればよい.
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6969 Accepted Submission(s): 2751
Problem Description
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.
Input
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).
Output
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.
Sample Input
2 1 10 2 3 15 5
Sample Output
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}.
AからBまでのnとの相互作用の数はどのくらいあるか.
1−Bにおけるnとの相互作用の個数と、1−(A−1)におけるnとの相互作用の個数を反発により求め、その後+1(1はいずれの数とも相互作用し、2回減算したので+1が必要)を減算すればよい.
#include
#include
using namespace std;
typedef long long ll;
#define maxn 100005
ll re[maxn];
ll k;
ll gcd(ll a,ll b) {
return b?gcd(b,a%b):a;
}
ll lcm(ll a,ll b)
{
return a/gcd(a,b)*b;
}
void primefactor(ll n)
{
k=0;
for(ll i=2;i*i<=n;i++)
{
if(n%i==0)
{
re[k++]=i;
while(n%i==0)
{
n/=i;
}
}
}
if(n!=1)
{
re[k++]=n;
}
}
int main()
{
ll n,t,a,b;
cin>>t;
ll ci=1;
while(t--)
{
cin>>a>>b>>n;
ll d=n;
primefactor(n);
ll ans1=0;
ll ans2=0;
for(ll i=1;i> j & 1)
{
cnt++;
temp=lcm(temp,re[j]);
}
}
if(cnt&1) ans1+=b/temp;
else ans1-=b/temp;
}
ans1=b-ans1;
for(ll i=1;i> j & 1)
{
cnt++;
temp=lcm(temp,re[j]);
}
}
if(cnt&1) ans2+=(a-1)/temp;
else ans2-=(a-1)/temp;
}
ans2=a-ans2;
cout<