Ultraviolet A 10780 again Prime? No time. messing around with quality factors

Find the maximum K so that m ^ K can be n! Division

M ^ K is to increase the number of each prime factor of m by K times, as long as N is obtained! It is enough to increase m by multiple times. Of course, the least increase factor is required here.

The amount of wood bottled water depends on the shortest plank.

The prefix is not cost-effective because N has 10000 and the first 10000 prime numbers have more than 1000 prime numbers.

Case is only 500, so we can simply calculate it every time.

#include<stdio.h>#include<string.h>#include<iostream>#include<math.h>#include<algorithm>#include<stdlib.h>#include<time.h>#include<vector>#include<map>using namespace std;struct node{    int x,t;}t,tzf[100001];vector<node> v[100001];vector<int> pri;int id[100011];bool vis[100011];void sieve(){    int m= (int)sqrt(200001+0.5);    memset(vis,0,sizeof(vis));    for(int i=2;i<=m;i++)    {        if(!vis[i])        {            for(int j=i*i;j<=10000;j+=i)                vis[j]=1;        }    }    for(int i=2;i<=10000;i++)    {        if(!vis[i])        {            pri.push_back(i);        }    }}void init(int n){    int id=n;    for(int i=0;pri[i]*pri[i]<=n&&i<pri.size();i++)    {        if(n%pri[i]==0)        {            t.x=pri[i];            t.t=1;            n/=pri[i];            while(n%pri[i]==0)            {                n/=pri[i];                t.t++;            }            v[id].push_back(t);        }    }    if(n>1)    {        t.x=n;        t.t=1;        v[id].push_back(t);    }}int sum[100005];int main(){    sieve();    for(int i=2;i<=10000;i++)        init(i);    int cas;    scanf("%d",&cas);    int ca=1;    while(cas--)    {        int n,m;        memset(sum,0,sizeof(sum));        scanf("%d%d",&n,&m);        printf("Case %d:\n",ca++);        int top=0;        for(int i=2;i*i<=n;i++)        {            if(n%i==0)            {                n/=i;                tzf[top].x=i;                tzf[top].t=1;                while(n%i==0)                {                    n/=i;                    tzf[top].t++;                }                top++;            }        }        if(n>1)        {            tzf[top].x=n;            tzf[top].t=1;            top++;        }        for(int i=2;i<=m;i++)        {            for(int j=0;j<v[i].size();j++)            {                int tmp=v[i][j].x;                int ttmp=v[i][j].t;                sum[tmp]+=ttmp;            }        }       int ans=0x7fffffff;        for(int i=0;i<top;i++)        {            int tmp=tzf[i].x;            int ttmp=tzf[i].t;            ans=min(ans,sum[tmp]/ttmp);        }        if(ans==0||ans==0x7fffffff) puts("Impossible to divide");        else cout<<ans<<endl;    }    return 0;}/*12342 102 10097 12*/