3643: Phi的反函数
Time Limit: 10 Sec Memory Limit: 64 MBDescription
Input
Output
Sample Input
4
Sample Output
5
HINT
Source
Tips:
自己也没怎么搞懂;
Code:
#include<cstdio> #include<algorithm> #include<cstring> #include<iostream> #define MAXN 200008 using namespace std; long long n,m,prime[MAXN],tot,flag[MAXN],ans; void init(){ for(long long i=2;i<=MAXN-8;i++){ if(!flag[i]){ flag[i]=1; prime[++tot]=i; } for(long long j=1;j<=tot&&i*prime[j]<=MAXN-8;j++){ flag[i*prime[j]]=1; if(i%prime[j]==0) break; } } } bool check(long long u){ for(int i=1;prime[i]*prime[i]<=u;i++){ if(u%prime[i]==0) return 0; } return 1; } void dfs(long long x,long long y,long long z){ if(z>=ans) return; if(y==1){ ans=min(ans,z); return; } if(y*y>n&&check(y+1)){ ans=min(ans,z*(y+1)); return; } for(int i=x;prime[i]-1<=y&&(prime[i]-1)*(prime[i]-1)<=n;i++) if(y%(prime[i]-1)==0){ long long xx=y /(prime[i]-1); long long yy=z*prime[i]; dfs(i+1,xx,yy); while(xx%prime[i]==0){ xx=xx/prime[i]; yy=yy*prime[i]; dfs(i+1,xx,yy); } } } int main(){ tot=0; ans=(long long)1<<31; init(); scanf("%lld",&n); dfs(1,n,1); if(ans==(long long)1<<31){ printf("-1"); }else{ printf("%lld",ans); } }