HDU 1796 How many integers can you find 解题报告（数论）

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3505    Accepted Submission(s): 982

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

 

    解题报告：0 - n 间能被集合中整除的数有多少个。暴力肯定不行，最大的n是2^31。

    思考一下，0 - n 间能被4整除的数有多少个？n / 4 个。

    那么 0 - n 间能被6整除的数有多少个？ n / 6 个。

    而 0 -  n 间能被{4, 6}整除的数应该是 n / 4 + n / 6 - n / lcm(4, 6) 个。因为有重复。

    重复此过程，集合中数字最多有10个，可以用数组存储每次lcm的值。

    AC代码如下：

#include <cstring>#include <cstdio>#include <algorithm>using namespace std;typedef long long LL;int array[21];int last[10000];int top;int gcd(int a, int b){return b == 0 ? a : gcd(b, a%b);}LL lcm(int a, int b){return (LL)a / gcd(a, b) * b;}int main(){#ifdef ACMfreopen("in.txt", "r", stdin);#endifint n, m;while (~scanf("%d%d", &n, &m)){n--;top = 0;for (int i = 0; i < m; i++){scanf("%d", array + i);if (array[i] == 0) m--, i--;}sort(array, array + m);int newM = 0;for (int i = 0; i < m; i++){bool ok = true;for (int j = 0; j < i; j++){if (array[i] % array[j] == 0){ok = false;break;}}if (ok){array[newM++] = array[i];}}m = newM;int res = 0;for (int i = 0; i < m; i++){int a = array[i];res += n / a;int newTop = top;last[newTop++] = a;for (int j = 0; j < top; j++){LL t;if (last[j] < 0)t = lcm(a, -last[j]);elset = -lcm(a, last[j]);if (t <= n){last[newTop++] = (int)t;res += n / (int)t;}}top = newTop;}printf("%d\n", res);}}

    上面这段不是太好看，但是偶然间发现这段代码是杭电上最快的= =，0MS

    标准点的容斥原理应该这么写：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int num[22];int gcd(int a, int b){    return b==0?a:gcd(b, a%b);}int lcm(int a, int b){    return a/gcd(a,b)*b;}int main(){    int n, m;    while(~scanf("%d%d", &n, &m))    {        n--;        for(int i=0; i<m; i++)        {            scanf("%d", num+i);            if(num[i]==0)                m--,i--;        }        int ans = 0;        for(int i=1;i<(1<<m);i++)        {            long long tmp = 1;            bool flag = false;            for(int j=0;j<m;j++) if(i&(1<<j))                tmp = lcm(tmp, num[j]), flag = !flag;            if(flag)                ans += n/tmp;            else                ans -= n/tmp;        }        printf("%d\n", ans);    }}

    因为多次重复lcm，所有代码的效率很慢。时效640MS。

    另外，本题和ZOJ 2836 近乎相同，将scanf m，n 的顺序交换即可。