SPOJ 6285. Another Game With Numbers

Little Chikoo likes to play with numbers. Often he plays the following game :
He chooses a number N and a set of positive integers.
He writes down all the numbers from 1 to N.
He chooses the first number (say x) from the set and cancels out all the multiples of x from 1 to N, including x.
He repeats step 3 for all the numbers from the set.

One day Little Chikoo was in a mood to play pranks. So his brother asked him to play the game with a certain challenge. He made the game a little harder and asked him to find out the number of integers which aren't cancelled after
he completes step 4. If he does that then Little Chikoo gets to play on his brother's Nintendo for one full day. Now Little Chikoo is in a hurry and wants to finish the job as soon as possible. He has asked for your help.

Input

The first line of the input contains N and K. (N <= 10^9, K <= 15)

Then K numbers follow all in a single line. All numbers are <= 100.

Output

The output file must contain the number of integers that aren't cancelled after he finishes step 4 of the game.

Example

Input:

10 3

2 4 5

Output:

4

(The numbers 1, 3, 7 and 9 weren't cancelled).

从1-N的数中，去掉K个数的倍数，求剩下的数的个数。用容斥原理。

最多15个数，于是用int m[15];来存。并且，用 int k 的前15个二进制位来表示是否取这个元素。

int AND(int k){//k为状态位，从右往左依次表示数组中的元素。计算同时是这些的倍数的数量。
if(CB[k]) return CB[k];//记忆化搜索，避免重复计算
long long D=1;
for(int i=0;i<15;i++){
if((k>>i)&1) D=lcm(D,m[i]);//取这些元素的最小公倍数
}
return CB[k]=N/D;返回1-N中，同时是选定元素的倍数的数的数量。
}

int OR(int k){//求1-N中，至少是选中的数中的某一个的倍数的数的数量。
int ANS=0;
for(int i=1;i<(1<<15)-1;i++){
int Nof1=0;//记录i中1的数量
for(int j=0;j<15;j++) if((i>>j)&1) Nof1++;
int T=Nof1&1;
if(!T) T=-1;//奇数为正，偶数为负。

ANS+=T*AND(i);
}
return ANS;
}

有了这俩函数，最后的输出语句：
printf("%d\n",N-OR( (1<<M)-1) );总数减去至少是某个倍数的数的个数。