HDU 1796How many integers can you find(简单容斥定理)
2013-10-14 09:56
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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): 3315 Accepted Submission(s): 937
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 2
2 3
Sample Output
7
Author
wangye
Source
2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up(4)
题目大意:很简单的题目,直接看意思就懂哈!
解题思路:容斥定理,加奇减偶,开始忘记求lcm了,囧!!而且开始还特判0的情况,题目中说的必须是除以,所以0不是一个解。。。开始竟然以为需要是因子就可以了。想通了之后直接先筛选一次,把0都筛选出去。
题目地址:How many integers can you find
AC代码:
#include<iostream> #include<cstring> #include<string> #include<cmath> #include<cstdio> using namespace std; __int64 sum; int n,m; int a[25]; int b[25]; int visi[25]; __int64 gcd(__int64 m,__int64 n) { __int64 tmp; while(n) { tmp=m%n; m=n; n=tmp; } return m; } __int64 lcm(__int64 m,__int64 n) { return m/gcd(m,n)*n; } void cal() { int flag=0,i; __int64 t=1; __int64 ans; for(i=0;i<m;i++) { if(visi[i]) { flag++; //记录用了多少个数 t=lcm(t,b[i]); } } ans=n/t; if(n%t==0) ans--; if(flag&1) sum+=ans; //加奇减偶 else sum-=ans; } int main() { int i,j,p; while(~scanf("%d%d",&n,&m)) { sum=0; for(i=0;i<m;i++) scanf("%d",&a[i]); int tt=0; // for(i=0;i<m;i++) { if(a[i]) //去掉0 b[tt++]=a[i]; } m=tt; p=1<<m; //p表示选取多少个数,组合数的状态 for(i=1;i<p;i++) { int tmp=i; for(j=0;j<m;j++) { visi[j]=tmp&1; tmp>>=1; } cal(); } printf("%I64d\n",sum); } return 0; } /* 12 2 2 3 12 3 2 3 0 12 4 2 3 2 0 */ //968MS
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