HDU 1796 How many integers can you find [容斥定理] 【组合数学】
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题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1796
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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): 7728 Accepted Submission(s): 2281
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
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题目大意:
求小于n的能够被集合中任意数字整除的数的个数
解题思路:
容斥原理入门题,
实现很好实现 可以状压搞,也可以dfs,但是发现dfs明显要块与状压啊。。
附本题代码
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状压枚举形式 655ms;
int a[100];int main(){ int n,m; while(~scanf("%d%d",&n,&m)){ for(int i=0;i<m;i++) { scanf("%d",&a[i]); if(a[i]==0) a[i]=INF; } sort(a,a+m); if(a[m-1]==INF) m--;/* printf("the numbers of %d (1<<m) = %d: \n",m,(1<<m)); for(int i=0;i<m;i++) printf("%d%c",a[i],(i==m-1)?'\n':' '); lalal;*/ n--; int ans = 0,num,sum; for(int i=1;i<(1<<m);i++){ num = 0, sum = 1; for(int j=0;j<m;j++){ if(i&(1<<j)){ sum=sum/__gcd(sum,a[j])*a[j]; num++; } } if(num&1) ans+=n/sum; else ans-=n/sum; //printf("%d %d\n",ans,sum); } printf("%d\n",ans); } return 0;}
dfs形式 202ms
int n,m;int a[100],ans;inline int gcd(int a,int b){return (!b)?a:gcd(b,a%b);}inline int lcm(int a,int b){return a/gcd(a,b)*b; }void dfs(int id,bool flag,int cnt){ cnt = lcm(a[id],cnt); if(flag ) ans += n/cnt; else ans -= n/cnt; for(int i=id+1;i<m;i++)dfs(i,!flag,cnt);}int main(){ while(~scanf("%d%d",&n,&m)){ n--; for(int i=0;i<m;i++) { scanf("%d",&a[i]); if(a[i]==0) a[i]=INF; } sort(a,a+m); if(a[m-1]==INF) m--; ans = 0; for(int i=0;i<m;i++) dfs(i,true,1); printf("%d\n",ans); } return 0;}
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