[原创]HDU 1796 How many integers can you find [容斥定理] 【组合数学】

2017-02-11 12:49:25 Tabris_ 阅读数：360

博客爬取于2020-06-14 22:41:40
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/54982226

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1796
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看详情——《IJCAI 2017 口碑商家客流量预测大赛》
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,

ans=nlcm{一个元素}−nlcm{两个元素}+nlcm{三个元素}−....nlcm{所有元素}ans = \dfrac{n}{lcm\{一个元素\} }-\dfrac{n}{lcm\{两个元素\} }+\dfrac{n}{lcm\{三个元素\} }-....\dfrac{n}{lcm\{所有元素\} }

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；

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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

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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;
}