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.
 

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.
 

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;
}




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