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HDU 4135:Co-prime (容斥原理)

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Co-prime Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4289Accepted Submission(s): 1698 Problem Description Given a number N, you are asked to count the number of integers between A and


Co-prime


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4289    Accepted Submission(s): 1698


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 <= 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: 10

思路:通常我们求1-n中与n互质的数的个数都是用欧拉函数! 但如果n比较大或者是求1~m中与n互质的数的个数等等问题,要想时间效率高的话还是用容斥原理!

容斥:先对n分解质因数,分别记录每个质因数, 那么所求区间内与某个质因数不互质的个数就是n / r(i),假设r(i)是r的某个质因子 假设只有三个质因子, 总的不互质的个数应该为p1+p2+p3-p1*p2-p1*p3-p2*p3+p1*p2*p3, 及容斥原理,可以转向百度百科查看相关内容 pi代表n/r(i),即与某个质因子不互质的数的个数 ,当有更多个质因子的时候, 可以用状态压缩解决,二进制位上是1表示这个质因子被取进去了。 如果有奇数个1,就相加,反之则相减。



AC代码:

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
typedef __int64 LL;
using namespace std;
vector<LL>prime;
void jprime(LL n) //分解质因子
{
for(LL i=2; i*i<=n; i++)
if(n%i==0)
{
prime.push_back(i); //质因子
while(n%i==0)
n/=i;
}
if(n>1)prime.push_back(n);
}

LL solve(LL a,LL b,LL n) //[a,b]中与n互质的数的个数
{
LL sum=0;
LL size=prime.size();
for(LL msk=1; msk<(1<<size); msk++) //共有2^size种组合情况
{
LL mult=1,bits=0;
for(LL i=0; i<size; i++) //枚举每一种情况
{
if(msk&(1<<i)) //如果在当前判断之内
{
++bits;
mult*=prime[i];
}
}
LL cur=b/mult-(a-1)/mult; //中间结果
if(bits&1)sum+=cur;
else sum-=cur;
}
return b-a-sum+1;
}
int main()
{
LL T;
scanf("%I64d",&T);
for(LL t=1; t<=T; t++)
{
LL a,b,n;
scanf("%I64d%I64d%I64d",&a,&b,&n);
prime.clear();
jprime(n);
printf("Case #%I64d: %I64d\n",t,solve(a,b,n));
}
return 0;
}
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