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

题目翻译：T组数据，每组给出A,B,N,求区间【A,B】里面有多少个数与N互质！

解题思路：求出所有的数字的与N不互质的情况，然后用总量减去！求不互质的情况，显然就是利用容斥原理，先求出N的质因子，然后去掉【A,B】中含有这些质因子的数，此时应该注意会重复去掉一些数字，记得加上即可，容斥原理的写法有好几种，感觉每种都很牛，比如用数组实现，DFS实现，以及用位运算实现！

数组实现：

#include<cstdio>
#define LL long long
LL p[10],k;//p数组用来保存n的质因子，LL型n不会超过10个
void getp(LL n){
    k=0;
    for(LL 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;//防止有比根号n大的质因子，k保存质因子个数
}
LL nop(LL m){
    LL i,j,que[10000],top=0,t,sum;
    que[top++]=-1;//队列数组保存n所有质因子任意不相同组合的乘积
    for(i=0;i<k;i++){
        t=top;//t保存当前que长度，方便下面的循环来使用
        for(j=0;j<t;j++){
            que[top++]=que[j]*p[i]*(-1);
        }//质因子的个数：奇加偶减，因此乘以-1来换号
    }
    for(i=1,sum=0;i<top;i++)//sum来累加所有个数
        sum+=m/que[i];
    return sum;
}
int main(){
    LL a,b,n;
    int t,i;
    scanf("%d",&t);
    for(i=1;i<=t;i++){
        scanf("%lld%lld%lld",&a,&b,&n);  //求1-m中多少个数字与n互质
        getp(n);//求n的质因子
        printf("Case #%d: %lld\n",i,b-nop(b)-(a-1-nop(a-1)));//总数减去
    }
    return 0;
}


 


DFS实现：

#include<stdio.h>
#include<math.h>
int p[10],top;
long long ansa,ansb,ans,a,b;
void DFS(int n,bool tag,long long num){
    if(n==top){
        if(tag==1){
            ansa-=a/num;
            ansb-=b/num;
        }
        else{
            ansa+=a/num;
            ansb+=b/num;
        }
        return;
    }
    DFS(n+1,tag,num);
    DFS(n+1,!tag,num*p[n]);
}
int main(){
    int i,j,n,T,k,cnt;
    cnt=1;
    scanf("%d",&T);
    while(T--){
        scanf("%I64d%I64d%d",&a,&b,&n);
        a--;
        ansa=ansb=0;
        top=0;
        for(i=2;i*i<=n;i++){
            if (n%i==0){
                while(n%i==0) n=n/i;
                p[top++]=i;
            }
        }
        if(n>1)
            p[top++]=n;
        DFS(0,0,1);
        printf("Case #%d: %I64d\n",cnt++,ansb-ansa);
    }
    return 0;
}

位运算

#include<cstdio>
#define LL long long
LL p[10],k;//p数组用来保存n的质因子，LL型n不会超过10个
void getp(LL n){
    k=0;
    for(LL 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;//防止有比根号n大的质因子，k保存质因子个数
}
LL nop(LL m){
    LL i,j,sum=0,flag,num;
    for(i=1;i<1<<k;i++){
        flag=0;
        num=1;
        for(j=0;j<k;j++)
            if(i&(1<<j))
                flag++,num*=p[j];
        if(flag&1) sum+=m/num;
        else       sum-=m/num;
    }
    return sum;
}
int main(){
    LL a,b,n;
    int t,i;
    scanf("%d",&t);
    for(i=1;i<=t;i++){
        scanf("%lld%lld%lld",&a,&b,&n);  //求1-m中多少个数字与n互质
        getp(n);//求n的质因子
        printf("Case #%d: %lld\n",i,b-nop(b)-(a-1-nop(a-1)));//总数减去
    }
    return 0;
}

 


 


 


 


神递归

#include<cstdio>
#define LL long long
LL p[10],k;//p数组用来保存n的质因子，LL型n不会超过10个
void getp(LL n){
    k=0;
    for(LL 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;//防止有比根号n大的质因子，k保存质因子个数
}
LL nop(LL m,LL t){
    LL i,sum=0;
    for(i=t;i<k;i++)
        sum+=m/p[i]-nop(m/p[i],i+1);
    return sum;
}
int main(){
    LL a,b,n;
    int t,i;
    scanf("%d",&t);
    for(i=1;i<=t;i++){
        scanf("%lld%lld%lld",&a,&b,&n);  //求1-m中多少个数字与n互质
        getp(n);//求n的质因子
        printf("Case #%d: %lld\n",i,b-nop(b,0)-(a-1-nop(a-1,0)));//总数减去
    }
    return 0;
}