Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3695 Accepted Submission(s): 1461
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}.
Source
题目大意：
给一个区间 [a,b] 求这个区间中 与 n 互素的个数。
解题思路：
这个问题可以转化为求 [1,b] 区间中与 n 互素的个数 ans1 减去 [1,a?1] 区间中与 n 互素的个
数 ans2 ,那么 [1,x] 区间中与 n 互素的个数怎么求呢，可以先将 n 进行素因子分解，然后用区间
x 除以 素因子，就得到了与 n 的 约数是那个素因子的个数，然后每次这样求一遍，但是发现有重
复的：举个例子 [1,10] 区间中与 6 互素的个数，应该是 10?(10/2+10/3) 但是这样多减去
了他们的最小公倍数 6 的情况，所以在加上 10/6 也就是：
10?(10/2+10/3?10/6)=3
这就用到了容斥原理，知道了这个，题目也就可以做了，先进行素因子分解，然后二进制枚举子集
进行容斥原理（奇加偶减）：
My Code：
/**
2016 - 08 - 08 上午
Author: ITAK
Motto:
今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 1e9+5;
const int MAXN = 1e6+5;
const int MOD = 1e9+7;
const double eps = 1e-7;
const double PI = acos(-1);
using namespace std;
bool prime[MAXN];
LL p[MAXN];
int k;
///素数筛选
void isprime()
{
k = 0;
memset(prime, 0, sizeof(prime));
for(LL i=2; i<MAXN; i++)
{
if(!prime[i])
{
p[k++] = i;
for(LL j=i*i; j<MAXN; j+=i)
prime[j] = 1;
}
}
}
LL fac[MAXN/100];
int cnt = 0;
void Dec(LL x)
{
cnt = 0;
for(int i=0; p[i]*p[i]<=x&&i<k; i++)
{
if(x%p[i]==0)
{
fac[cnt++] = p[i];
while(x%p[i]==0)
x /= p[i];
}
}
if(x > 1)
fac[cnt++] = x;
}
LL Solve(LL x)
{
LL ret = 0;
for(LL i=1; i<(1LL<<cnt); i++)
{
int sum = 0, tmp = 1;
for(int j=0; j<cnt; j++)
{
if(i & (1LL<<j))
{
sum++;
tmp *= fac[j];
}
}
if(sum & 1)
ret += x/tmp;
else
ret -= x/tmp;
}
return ret;
}
int main()
{
isprime();
int T;
LL N, A, B;
scanf("%d",&T);
for(int cas=1; cas<=T; cas++)
{
scanf("%I64d%I64d%I64d",&A,&B,&N);
Dec(N);
printf("Case #%d: %I64d\n",cas,B-Solve(B)-A+1+Solve(A-1));
}
return 0;
}