[Solution] bzoj4807 problem of high precision linear combination of several vehicle screening

Topic links: https://www.lydsy.com/JudgeOnline/problem.php?id=4807

The label is mathematics teacher.

I thought hard to get.

The results found that the maximum number of put this restriction.

The answer is natural to think such a diagonal arrangement.

NATURAL i.e., C (max (n, m), min (n, m))

Mainly want to keep 50.

It must be high precision.

Then in order to avoid high-precision division.

Prime factor decomposition.

About vertical minutes then high-precision multiplication.

code show as below:

/**************************************************************
    Problem: 4807
    User: ChrisK
    Language: C++
    Result: Accepted
    Time:304 ms
    Memory:13012 kb
****************************************************************/
 
#include<bits/stdc++.h>
using namespace std;
const int maxn=1000010;
int n,m,tot,l=50;
int p[maxn],q[maxn];
int c[maxn];
struct node{
    int s[60];
    node operator * (const int &b) const{
        node q;
        for(int i=1;i<=50;i++)
            q.s[i]=0;
        for(int i=1;i<=50;i++){
            q.s[i]+=s[i]*b;
            q.s[i+1]+=q.s[i]/10;
            q.s[i]%=10;
        }
        return q;
    }
}ans;
void chai(int x,int y){
    while(x!=1){
        c[p[x]]+=y,x/=p[x];   
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    if(n<m) swap(n,m);
    for(int i=2;i<=n;i++){
        if(!p[i])
            p[i]=i,q[++tot]=i;
        for(int j=1;j<=tot&&i*q[j]<=n;j++){
            p[i*q[j]]=q[j];
            if(i%q[j]==0) break;
        }
    }
    for(int i=1;i<=n-m;i++)
        chai(i,-1);
    for(int i=m+1;i<=n;i++)
        chai(i,1);
    ans.s[1]=1;
    for(int i=2;i<=n;i++)
        for(int j=1;j<=c[i];j++)
            ans=ans*i;
    while(ans.s[l]==0)
        l--;
    for(int i=l;i>=1;i--)
        printf("%d",ans.s[i]);
    puts("");
 
    return 0;
}