【HDU】 1923 Exploding CPU
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Exploding CPU
题目链接
- Exploding CPU
题目大意
规定一种数 explosive number,假设x是这种数,则x需要满足
现在给你一个范围,要你求出这个范围内的所有explosive number。
题解
这一题的关键是要推出A和B的范围,因为p最多只有10个,最小有3个,我们可以确定A和B的范围然后直接枚举质因子进行打表。而经过推算A和B有不是很大,所以此题采用打表。
POJ上有一段很经典的推理,这里给出链接:A和B范围的推算
代码
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define LL long longusing namespace std;int p[60000],h,num[1005],t,T,low,high;bool vis[600005];void deal(int high){ h=0; p[h++]=2; for (int i=2;i<=high;i+=2) vis[i]=1; for (int i=3;i<=high;i+=2) if (!vis[i]) { p[h++]=i; for (int j=i;j<=high;j+=i) vis[j]=1; }}int ser(int low,int high,LL x,int p[]){ int l=low,r=high,mid; while(l!=r) { mid=(l+r)>>1; if (x<=p[mid]) r=mid; else l=mid+1; } return l;}void solve(){ LL pri[15]; t=0; for (int a=1;a<=600;a++) for (int b=-700;b<=700;b++) { int i; memset(pri,0,sizeof(pri)); pri[0]=1; for (i=1;i<=10;i++) { pri[i]=pri[i-1]*a+b; if (p[ser(0,h-1,pri[i],p)]!=pri[i]) break; } if (i<4) continue; LL tmp=pri[1]*pri[2]*pri[3]; if (tmp>2000000000) continue; num[t++]=tmp; for (int j=4;j<i;j++) { tmp*=pri[j]; if (tmp<=2000000000) num[t++]=tmp; else break; } } sort(num,num+t);}int main(){ deal(600000); solve(); scanf("%d",&T); while(T--) { int l,r; scanf("%d%d",&low,&high); l=ser(0,t-1,low,num); r=ser(0,t-1,high,num); if (num[r]>high) r--; printf("%d\n",r-l+1); } return 0;}
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