uva10837 A Research Problem

A mad researcher was trying to get fund for his research project but

it is a pity that after a year he was able to collect 500 $ only. So

all his research work has gone to ashtray. The mad researcher now

wants his revenge, so he wants you to solve his un nished research

problem within a very limited time. You will be happy to know that his

research is related to Eulers phi function. Euler’s phi (or totient)

function of a positive integer n is the number of integers in f 1 ; 2

; 3 ; : : : ; n g which are relatively prime to n . This is usually

denoted as φ ( n ). The table below shows the value of phi function

for rst few numbers. integer n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

φ ( n ) 1 1 2 2 4 2 6 4 6 4 10 4 12 6 8 8 Given the value of n , it is

very easy to nd the value of φ ( n ) using the formula below: φ ( n )

= n ∏ p j n (1

根据phi(n)=n * (p1-1)/p1 * (p2-1)/p2 * … * (pk-1)/pk，所有n的质因数-1一定是m的因数。找到sqrt(m)以内的所有质数，对每个质数是否取以及次数进行搜索，最后再判断一下大于sqrt(m)的因数【最多只有一个，次数最高为1】。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define LL long long
const int maxp=100000,oo=0x3f3f3f3f;
int p[100000],tot,a[100000],cnt,m,ans;
bool have[100010];
bool is(int x)
{
int i;
int m=sqrt(x+0.1);
for (i=1;i<=tot&&p[i]<=m;i++)
if (x%p[i]==0)
return 0;
return 1;
}
void make()
{
int i,j;
for (i=2;i<=maxp;i++)
{
if (!have[i]) p[++tot]=i;
for (j=1;j<=tot&&(LL)i*p[j]<=maxp;j++)
{
have[i*p[j]]=1;
if (i%p[j]==0) break;
}
}
}
void init()
{
int i;
cnt=0;
for (i=1;i<=tot;i++)
if (m%(p[i]-1)==0)
a[++cnt]=p[i];
}
void dfs(int k,LL n,LL phi)
{
if (phi==m) ans=min(ans,(int)n);
if (k==cnt+1)
{
if (m%phi==0&&m/phi>a[cnt]&&is(m/phi+1))
ans=min(ans,(int)(n*(m/phi+1)));
return;
}
dfs(k+1,n,phi);
phi*=(a[k]-1);
for (n*=a[k];n<=ans;n*=a[k],phi*=a[k])
dfs(k+1,n,phi);
}
int main()
{
int K=0;
make();
while (scanf("%d",&m)&&m)
{
printf("Case %d: %d ",++K,m);
init();
ans=oo;
dfs(1,1,1);
printf("%d\n",ans);
}
}