标签：algorithm span contain any rom line str color math
Given a list of integers (A1, A2, ..., An), and a positive integer M, please find the number of positive integers that are not greater than M and dividable by any integer from the given list.
Input
The input contains several test cases.
For each test case, there are two lines. The first line contains N (1 <= N <= 10) and M (1 <= M <= 200000000), and the second line contains A1, A2, ..., An(1 <= Ai <= 10, for i = 1, 2, ..., N).
Output
For each test case in the input, output the result in a single line.
Sample Input
3 2
2 3 7
3 6
2 3 7
Sample Output
1
4
分析：1~m中至少能被列表里其中一个数整除的数的个数；
容斥即可，注意列表中的数不一定互质，所以要求lcm；
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <bitset> #include <map> #include <queue> #include <stack> #include <vector> #define rep(i,m,n) for(i=m;i<=n;i++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define sys system("pause") const int maxn=1e5+10; using namespace std; inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} inline void umax(ll &p,ll q){if(p<q)p=q;} inline void umin(ll &p,ll q){if(p>q)p=q;} inline ll read() { ll x=0;int f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } int n,m,k,t,fac[10]; int main() { int i,j; while(~scanf("%d%d",&n,&m)) { rep(i,0,n-1)scanf("%d",&fac[i]); int ret=0; rep(i,1,(1<<n)-1) { int now=1,cnt=0; rep(j,0,n-1) { if(i&(1<<j)) { cnt++; now=now*fac[j]/gcd(now,fac[j]); } } if(cnt&1)ret+=m/now; else ret-=m/now; } printf("%d\n",ret); } return 0; }
标签：algorithm span contain any rom line str color math
原文地址：http://www.cnblogs.com/dyzll/p/6360097.html
