Time Limit: 500/200 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 1281 Accepted Submission(s): 319
题解:当type是0的时候是反素数模版,不过要剪枝,否则超时;当type是1
的时候是因子个数是x-k个;用俩重循环;
代码:
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
const double PI=acos(-1.0);
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define T_T while(T--)
#define P_ printf(" ")
int prim[16]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
typedef unsigned long long uLL;
const int MAXN=60010;
const uLL INF=(uLL)~0;
uLL ans;
int k;
int dp[MAXN];
void initial(){
for(int i=0;i<MAXN;i++)dp[i]=i;
for(int i=1;i<MAXN;i++){
for(int j=i;j<MAXN;j+=i)dp[j]--;
if(dp[dp[i]]==0)//如果没被使用就让dp[dp[i]]=i; **
dp[dp[i]]=i;
dp[i]=0;//当前值已经没用了,就为0 **
}
}
void dfs(int pos,uLL v,int num){
if(num==k&&ans>v)ans=v;
for(int i=1;i<=62;i++){
if(num*(i+1)>k||ans/prim[pos]<v)break;//剪枝;
v*=prim[pos];
if(k%(num*(i+1))==0)dfs(pos+1,v,num*(i+1));//剪枝
}
}
int main(){
int T,kase=0;
SI(T);
initial();
T_T{
int type;
SI(type);SI(k);
ans=INF;
if(type)ans=dp[k];
else{
dfs(0,1,1);
}
printf("Case %d: ",++kase);
if(ans==0)puts("Illegal");
else if(ans==INF)puts("INF");
else printf("%llu\n",ans);
}
return 0;
}
原文:http://www.cnblogs.com/handsomecui/p/5017882.html