HDU

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7953    Accepted Submission(s): 2364

Problem Description

  Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.

 

Input

  There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.

 

Output

  For each case, output the number.

 

Sample Input

12 22 3

 

Sample Output

7

 

Author

wangye

 

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

 

Recommend

wangye

 

题目大意：给定n和一个大小为m的集合，集合元素为非负整数。为1...n内能被集合里至少一个数整除的数字个数。n<=2^31,m<=10

trick：给的集合里面可能有0，所有取模，gcd，lcm都要注意有没有0，有0要提出来。。。 

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;typedef long long ll;ll a[15];int main(){    ll n, m;    while(~scanf("%lld%lld", &n, &m))    {        int cnt = 0, x;        for(int i = 1; i <= m; i++)        {            scanf("%d", &x);            if(x!=0) a[cnt++] = x;        }        ll ans = 0;        for(int i = 1; i < (1<<cnt); i++)  //容斥代码， 奇加偶减。。        {            int sum = 0;            for(int j = i; j > 0; j >>= 1)                sum += j&1;            ll lcm = 1;            for(int j = 0; j < cnt; j++)            {                if((1<<j)&i)                    lcm = lcm/__gcd(a[j], lcm)*a[j];            }            if(sum%2) ans += (n-1)/lcm;            else ans -= (n-1)/lcm;        }        printf("%lld\n", ans);    }    return 0;}

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4683    Accepted Submission(s): 1867

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

 

Recommend

lcy

 

思路i转自：http://www.cnblogs.com/jiangjing/archive/2013/06/03/3115470.html

题意：就是让你求(a,b)区间于n互质的数的个数.

分析：我们可以先转化下：用(1,b)区间与n互质的数的个数减去(1,a-1)区间与n互质的数的个数,那么现在就转化成求(1,m)区间于n互质的数的个数,如果要求的是(1,n)区间与n互质的数的个数的话，我们直接求出n的欧拉函数值即可，可是这里是行不通的!我们不妨换一种思路：就是求出(1,m)区间与n不互质的数的个数，假设为num，那么我们的答案就是：m-num!现在的关键就是：怎样用一种最快的方法求出(1,m)区间与n不互质的数的个数？方法实现：我们先求出n的质因子(因为任何一个数都可以分解成若干个质数相乘的),如何尽快地求出n的质因子呢？我们这里又涉及两个好的算法了!第一个：用于每次只能求出一个数的质因子,适用于题目中给的n的个数不是很多，但是n又特别大的;(http://www.cnblogs.com/jiangjing/archive/2013/06/03/3115399.html)第二个：一次求出1~n的所有数的质因子,适用于题目中给的n个数比较多的，但是n不是很大的。(http://www.cnblogs.com/jiangjing/archive/2013/06/01/3112035.html)本题适用第一个算法！举一组实例吧:假设m=12,n=30.

第一步：求出n的质因子：2,3,5；

第二步：(1,m)中是n的因子的倍数当然就不互质了(2,4,6,8,10)->n/2  6个,(3,6,9,12)->n/3  4个,(5,10)->n/5  2个。

如果是粗心的同学就把它们全部加起来就是：6+4+2=12个了，那你就大错特错了,里面明显出现了重复的,我们现在要处理的就是如何去掉那些重复的了！

第三步：这里就需要用到容斥原理了，公式就是：n/2+n/3+n/5-n/(2*3)-n/(2*5)-n/(3*5)+n/(2*3*5).

第四步:我们该如何实现呢？我在网上看到有几种实现方法：dfs(深搜)，队列数组，位运算三种方法都可以！上述公式有一个特点：n除以奇数个数相乘的时候是加，n除以偶数个数相乘的时候是减。我这里就写下用队列数组如何实现吧：我们可以把第一个元素设为-1然后具体看代码如何实现吧！

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>using namespace std;typedef long long ll;const int maxn = 1e3;ll prime[maxn], l, r, n;int cnt;void get_prime(ll num){    cnt = 0;    for(ll i = 2; i*i <= num; i++)    {        if(!num) break;        if(num%i == 0)        {            prime[cnt++] = i;            while(num%i == 0) num /= i;        }    }    if(num > 1) prime[cnt++] = num;}ll rep(ll num){    ll ans = 0;    for(int i = 1; i < (1<<cnt); i++)    {        int sum = 0;        ll val = 1;        for(int j = 0; j < cnt; j++)        {            if(i & (1<<j))                val *= prime[j], sum++;        }        if(sum&1) ans += num/val;        else ans -= num/val;    }    return num-ans;}int main(){    int t, ca = 1;    scanf("%d", &t);    while(t--)    {        scanf("%lld%lld%lld", &l, &r, &n);        memset(prime, 0, sizeof(prime));        get_prime(n);        printf("Case #%d: %lld\n",ca++, rep(r)-rep(l-1));    }    return 0;}