codeforce 1017 F. The Neutral Zone

F. The Neutral Zone

time limit per test

5 seconds

memory limit per test

16 megabytes

input

standard input

output

standard output

Notice: unusual memory limit!

After the war, destroyed cities in the neutral zone were restored. And children went back to school.

The war changed the world, as well as education. In those hard days, a new math concept was created.

As we all know, logarithm function can be described as:log(pa11pa22...pa2k)=a1logp1+a2logp2+...+aklogpklog⁡(p1a1p2a2...pka2)=a1log⁡p1+a2log⁡p2+...+aklog⁡pkWhere pa11pa22...pa2kp1a1p2a2...pka2 is the prime factorization of a integer. A problem is that the function uses itself in the definition. That is why it is hard to calculate.

So, the mathematicians from the neutral zone invented this:exlogf(pa11pa22...pa2k)=a1f(p1)+a2f(p2)+...+akf(pk)exlogf(p1a1p2a2...pka2)=a1f(p1)+a2f(p2)+...+akf(pk)

Notice that exlogf(1)exlogf(1) is always equal to 00.

This concept for any function ff was too hard for children. So teachers told them that ff can only be a polynomial of degree no more than 33in daily uses (i.e., f(x)=Ax3+Bx2+Cx+Df(x)=Ax3+Bx2+Cx+D).

"Class is over! Don't forget to do your homework!" Here it is:n∑i=1exlogf(i)∑i=1nexlogf(i)

Help children to do their homework. Since the value can be very big, you need to find the answer modulo 232232.

Input

The only line contains five integers nn, AA, BB, CC, and DD (1≤n≤3⋅1081≤n≤3⋅108, 0≤A,B,C,D≤1060≤A,B,C,D≤106).

Output

Print the answer modulo 232232.

思路：
每个素数p对答案的贡献是 n/(p^1)+n/(p^2)+n/(p^3)....
即 f(p)*{n/(p^1)+n/(p^2)+n/(p^3)....} p取遍2~n中的所有素数。
数据有3e8,只用bitset优化还是会超限，因为不能被2,3整除的数除以3
等于它排列的下标值：如前20个数中不能被2,3整除的有：
a[1]=5,a[2]=7,a[3]=11,a[4]=13,a[5]=17,a[6]=19。
有a[1]/3=1,a[2]/3=2,a[3]/3=3,a[4]/3=4,a[5]/3=5,a[6]/3=6。
因此在用数组标记素数的时候可以把是2或3的倍数的数剔除掉。
这样就可以开一个1e8的bitset。
代码：

#include<bits/stdc++.h>
using namespace std;
#define uint unsigned int
const int maxn=1e8+10;
bitset<maxn>G;
uint n,A,B,C,D;
uint get_ans(uint x)
{
    uint ans=0;
    for(uint i=1;i<=n/x;i*=x) ans+=n/(i*x);
    return ans*(A*x*x*x+B*x*x+C*x+D);
}
int main()
{
    scanf("%u%u%u%u%u",&n,&A,&B,&C,&D);
    uint ans=get_ans(2)+get_ans(3);
    for(uint i=5;i<=n;i++)
    {
        if(i%2==0||i%3==0) continue;
        if(!G[i/3])
        {
            ans+=get_ans(i);
            for(int j=i;j<=n;j+=i)
            {
                if(j%2==0||j%3==0) continue;
                G[j/3]=1;
            }
        }
    }
    printf("%u\n",ans);
    return 0;
}

codeforce 1017 F. The Neutral Zone

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