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