Today is Yukari's n-th birthday. Ran and Chen hold a celebration party for her. Now comes the most important part, birthday cake! But it's a big challenge for them to place n candles on the top of the cake. As Yukari has lived for such a long long time. Though she herself insists that she is a 17-year-old girl.
To make the birthday cake look more beautiful, Ran and Chen decide to place them like r ≥ 1 concentric circles. They place ki candles equidistantly on the i-th circle, where k ≥ 2, 1 ≤ i ≤ r. And it's optional to place at most one candle at the center of the cake. In case that there are a lot of different pairs of r and k satisfying these restrictions, they want to minimize r × k. If there is still a tie, minimize r.
There are about 10,000 test cases. Process to the end of file.
Each test consists of only an integer 18 ≤ n ≤ 1012.
For each test case, output r and k.
18 111 1111
1 17 2 10 3 10
大坑题 ,枚举r 二分k,这样就可以了, 但是要注意 ,会出现__int64爆掉的情况,这样的话,那就要加很多的判断了!
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; __int64 n; int fccos(__int64 x,int ii) { __int64 sum=0,kk=1; int i; for(i=1;i<=ii;i++) { if(kk>n/x) return 0; kk*=x; if(kk>n||kk<0) return 0; sum+=kk; // printf(" %I64d ",sum); if(sum>n||sum<0) return 0; } //printf("%I64d %I64d ii%d\n",sum,x,ii); if(sum==n||sum==n-1) return 2; if(sum>n) return 0; if(sum<n-1) return 1; } __int64 twocut(int x) { __int64 s,e,mid; s=1;e=n; while(s<=e) { mid=(s+e)>>1; int temp=fccos(mid,x); if(temp==2) { return mid; } else if(temp==1) s=mid+1; else if(temp==0) e=mid-1; // printf("%d ",mid); } return -1; } int main() { int i,num; __int64 maxx,ansr,ansk; while(scanf("%I64d",&n)!=EOF) { maxx=-1,ansr=-1,ansk=-1; //num=log((double)n)/log(2.0)+1; //printf("num%d\n",num); for(i=1;i<=60;i++) { __int64 k=twocut(i); if(k<0) continue; __int64 temp=i*k; if(maxx==-1) { maxx=temp; ansr=i; ansk=k; } else if(temp<maxx) { maxx=temp; ansr=i; ansk=k; } else if(temp==maxx&&i<ansr) { maxx=temp; ansr=i; ansk=k; } } printf("%I64d %I64d\n",ansr,ansk); } return 0; }