hdoj 4135 Co-prime 【容斥原理】
Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1882 Accepted Submission(s): 722
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.
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 <= 10
15) and (1 <=N <= 10
9).
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: 10HintIn the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.
容斥原理：队列实现
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<stack>
#include<algorithm>
#include<iostream>
using namespace std;
long long p[100],k;
void getp(long long n)
{
long long i;
k=0;
for(i=2;i*i<=n;i++)
{
if(n%i==0)
p[k++]=i;
while(n%i==0)
n/=i;
}
if(n>1)
p[k++]=n;
}
long long nop(long long m)
{
long long que[100000],i,j,sum,top=0,t;
que[top++]=-1;
for(i=0;i<k;i++)
{
t=top;
for(j=0;j<t;j++)
{
que[top++]=que[j]*p[i]*(-1);
}
}
for(i=1,sum=0;i<top;i++)
sum+=m/que[i];
return sum;
}
int main()
{
long long a,b,n;
int t,j=1;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld%lld",&a,&b,&n);
getp(n);
printf("Case #%d: %lld\n",j++,b-nop(b)-(a-1-nop(a-1)));
}
return 0;
}
更新2015.6.14
位运算：
#include <cstdio>
#include <cstring>
#define LL long long
LL p[100], k;
void getp(LL n)
{
LL i;
k = 0;
for(i = 2; i*i <= n; i++)
{
if(n % i == 0)
{
p[k++] = i;
while(n % i == 0)
n /= i;
}
}
if(n > 1) p[k++] = n;
}
LL nop(LL m)
{
LL i, j;
LL use, res;
LL sum = 0;
for(i = 1; i < 1<<k; i++)
{
use = 0;
res = 1;
for(j = 0; j < k; j++)
{
if(i & (1<<j))
{
use++;
res *= p[j];
}
}
if(use & 1) sum += m / res;
else sum -= m / res;
}
return sum;
}
int main()
{
int t;
int j = 1;
LL a, b, n;
scanf("%d", &t);
while(t--)
{
scanf("%lld%lld%lld", &a, &b, &n);
getp(n);
printf("Case #%d: %lld\n", j++, b-nop(b)-(a-1-nop(a-1)));
}
return 0;
}