正文
hdu 1796(容斥原理+状态压缩)
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How many integers can you find
Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6630 Accepted Submission(s): 1913
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.
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.
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 2
2 3
2 3
Sample Output
7
题意:在 1-(n-1) 中能够被输入的数字整除的数字的数量。
思路:容斥原理+枚举状态,碰到奇数加上(n-1)/lcm(a,b,c..) 碰到偶数减(n-1)/lcm(a,b,c...) 注意0不能取，发现本人一直不是很会用深搜，所以还是用状压了 = =
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <math.h>
using namespace std;
int gcd(int a,int b){
return b==?a:gcd(b,a%b);
}
int lcm(int a,int b){
return a/gcd(a,b)*b;
}
int main()
{
int n,m,a[];
while(scanf("%d%d",&n,&m)!=EOF){
int id = ,num;
for(int i=;i<m;i++){
scanf("%d",&num);
if(num!=) a[id++] = num;
}
int ans = ;
for(int i=;i<(<<id);i++){
int l=,cnt=;
for(int j=;j<id;j++){
if((i>>j)&){
cnt++;
l = lcm(l,a[j]);
}
}
if(cnt&){
ans+=(n-)/l; ///不包括自身所以n-1
}else{
ans-=(n-)/l;
}
}
printf("%d\n",ans);
}
return ;
}
