hdu4430 Yukari's Birthday 2012 ACM_ICPC Asia ChangChun Regional Contest problem K
题意：有个人过生日在蛋糕上插蜡烛，一圈一圈的插成同心圆，要求每一圈从里到外正好依次插k的i（1~r）次幂个蜡烛，最中间圆心可插可不插，给一个n（18~1w亿）,输出k和r。并且要求k*r尽可能小，同时r尽可能小。数据一共1w组左右。
思路：看到有1w组n，并且n是10的12次方，那明显要用2分才能做了。题目时间给的很充足，所以先打个表。把小于1w亿的k的r次幂都打到表里（1次幂就不要打了，会超内存）注意底数
只要取到10的6次幂就行，最后打出来的表大小101w出头，然后有个很重要的工作是去重（因为这个WA了几次）把相邻的值一样的结点比较一下，取最优值留下。然后就2分找就行了，找到一个点注意再看看他前面的那个值能不能+1等于n，再比较一下。这样除了打表，时间复杂度是T*log(100w)总共20w，基本可以忽略了。
Yukari's Birthday
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4181 Accepted Submission(s): 946
Problem Description
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 k i 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.
To make the birthday cake look more beautiful, Ran and Chen decide to place them like r ≥ 1 concentric circles. They place k i 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.
Input
There are about 10,000 test cases. Process to the end of file.
Each test consists of only an integer 18 ≤ n ≤ 10 12.
Each test consists of only an integer 18 ≤ n ≤ 10 12.
Output
For each test case, output r and k.
Sample Input
18 111 1111
Sample Output
1 17 2 10 3 10
Source
2012 Asia ChangChun Regional Contest
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <stack>
using namespace std;
typedef long long LL;
const int INF=0x7fffffff;
const int MAX_N=10000;
LL n;
struct node{
long long r,k;
long long v;
friend bool operator<(const node &a,const node &b){
return a.v<b.v;
}
};
node A[1020000];
bool cmp(const node &a,const node &b){
return a.v<b.v;
}
node cmpnode(node a,node b){
if(a.r*a.k<b.r*b.k)return a;
else if(a.r*a.k==b.r*b.k){
if(a.r<b.r)return a;
}
return b;
}
int main(){
int num=0;
memset(A,0,sizeof(A));
for(long long i=2;i<1000000;i++){
long long sum=i;
long long t=1;
for(long long j=i*i;sum<=1000000000000;j*=i){
t++;
sum+=j;
if(sum<=1000000000000){//&&(t*i<sum-1)这个写不写都行，不存在r*k<n-1的情况
A[num].r=t;
A[num].k=i;
A[num].v=sum;
num++;
// A[num].r=t;
// A[num].k=i;
// A[num].v=sum+1;
// num++;
}
}
}
// cout<<num<<endl;
sort(A,A+num,cmp);
// cout<<num<<endl;
// for(int i=0;i<200;i++){
// printf("%lld\n",A[i].v);
// }
//去重，同等数字只保留最优值
int newnum=0;
for(int i=0;i<num;i++){
// if(A[i].v==A[i+1].v&&A[i+1].v==A[i+2].v)cout<<"fuck";
//通过此句判断出不存在3连等的情况
if(A[i].v==A[i+1].v){
// cout<<A[i].v<<"fuck"<<endl;
A[newnum++]=cmpnode(A[i],A[i+1]);
i++;
}
else A[newnum++]=A[i];
}
// cout<<newnum<<endl;
node f;node c;
f.v=0;
node *o=lower_bound(A,A+num,f);
while(scanf("%lld",&n)!=EOF){
c.v=n;
//node l=*lower_bound(A,A+num,c);
int l=lower_bound(A,A+num,c)-o;
int r=upper_bound(A,A+num,c)-o;
// cout<<l<<r<<endl;
// node m
if(l==r){
if(A[l-1].v+1==n){
if(A[l-1].r*A[l-1].k<n-1){
printf("%lld %lld\n",A[l-1].r,A[l-1].k);
continue;
}
}
}
else{
node m=A[l];
if(A[l-1].v+1==n){
if(A[l-1].r*A[l-1].k<n-1){
m=cmpnode(A[l-1],m);
}
}
if(m.r*m.k<n-1){
printf("%lld %lld\n",m.r,m.k);
continue;
}
}
printf("1 %lld\n",n-1);
}
return 0;
}