Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7178 Accepted Submission(s): 2836
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}.
PS：这个题如果用暴力直接跑a，b。肯定会超时的，所以这里我们可以用到容斥定理，我们求a，b里与n互质的个数，可以求这个范围内与n不互质的数的个数，再用总个数减去这个个数，就是答案。首先我们算1~b里与n不互质的个数，再算1~a里与n不互质的个数。两数相减，就是a~b中与n不互质的个数了。
举一组实例吧:假设m=12,n=30.。
第一步：求出n的质因子：2,3,5；
第二步：(1,m)中是n的因子的倍数当然就不互质了(2,4,6,8,10)=n/2 6个,(3,6,9,12)=n/3 4个,(5,10)=n/5 2个。6+4+2=12个了，里面明显出现了重复的,我们现在要处理的就是如何去掉那些重复的了！
第三步：这里就需要用到容斥原理了，公式就是：n/2+n/3+n/5-n/(2*3)-n/(2*5)-n/(3*5)+n/(2*3*5).
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<cmath>
const int maxn=1e4+10;
const int mod=1e9+7;
const int inf=1e9+10;
#define me(a,b) memset(a,b,sizeof(a))
#define PI 3.1415926
#define e exp(1.0)
typedef long long ll;
using namespace std;
ll a[500],l;
void ola(ll n)
{
l=0;
for(ll i=2;i*i<=n;i++)
{
if(n%i==0)
a[l++]=i;///n的因子
while(n%i==0)
n/=i;
}
if(n>1)
a[l++]=n;
}
ll paici(ll m)
{
ll temp[maxn],t=0,sum=0;me(temp,0);
temp[t++]=-1;
for(int i=0;i<l;i++)
{
int k=t;
for(int j=0;j<k;j++)
temp[t++]=temp[j]*a[i]*(-1);///这步可以自己单步调试下，应该就能理解。
}
for(int i=1;i<t;i++)
sum+=m/temp[i];
return sum;
}
int main()
{
int t,Case=1;cin>>t;
while(t--)
{
ll a,b,n;cin>>a>>b>>n;
ola(n);///找因子
ll s=b-paici(b)-(a-1-paici(a-1));
cout<<"Case #"<<Case++<<": "<<s<<endl;
}
return 0;
}