[容斥原理] hdu 4135 Co-prime

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4135

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1176    Accepted Submission(s): 427

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

21 10 23 15 5

 

Sample Output

Case #1: 5Case #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

The Third Lebanese Collegiate Programming Contest

 

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lcy  

题目意思：
给一个a,b,n求在[a,b]内有多少个与n互质的数。

解题思路：

简单的容斥原理。

反面思考，先求出与n不互质的，也就是包含n的质因数的。然后就可以想到先把n分解质因数。然后可以想到先预处理1000000内的质数。

然后求出1~b满足要求的个数，减去1~a-1满足要求的个数，答案即为最后结果。

代码：

//#include<CSpreadSheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define N 1000000int prime[130000],cnt;bool isprime[N+10];int pp[110000],cnt0;ll a,b,n,ans1,ans2;void init() //预处理出1~1000000内的质数{    cnt=0;    for(int i=1;i<=N;i++)        isprime[i]=true;    for(int i=2;i<=N;i++)    {        if(!isprime[i])            continue;        prime[++cnt]=i;        for(int j=i*2;j<=N;j+=i)            isprime[j]=false;    }    //printf("cnt:%d\n",cnt);}void Cal(ll cur) //分解出cur的质因数{    cnt0=0;    for(int i=1;prime[i]*prime[i]<=cur;i++)    {        if(cur%prime[i]==0)        {            pp[++cnt0]=prime[i];            while(!(cur%prime[i]))                cur/=prime[i];        }    }    if(cur!=1)        pp[++cnt0]=cur;}void dfs(ll hav,int cur,int num) //容斥原理求出互质的{    if(cur>cnt0||(hav>a&&hav>b))        return ;    for(int i=cur;i<=cnt0;i++)    {        ll temp=hav*pp[i];        if(num&1)         {            ans1-=b/temp;            ans2-=a/temp;        }        else        {            ans1+=b/temp;            ans2+=a/temp;        }        dfs(temp,i+1,num+1);    }}int main(){   //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   init();   //system("pause");   int t;   scanf("%d",&t);   for(int ca=1;ca<=t;ca++)   {       scanf("%I64d%I64d%I64d",&a,&b,&n);       Cal(n);       //printf("cnt0:%d\n",cnt0);       a--;       ans1=b,ans2=a;       for(int i=1;i<=cnt0;i++)       {            ans1-=b/pp[i];            ans2-=a/pp[i];            dfs(pp[i],i+1,2);       }       printf("Case #%d: %I64d\n",ca,ans1-ans2);   }   return 0;}