hdu 4430 枚举+二分 范围运算失误

题目：

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5683    Accepted Submission(s): 1370

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 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.

 

Input

There are about 10,000 test cases. Process to the end of file.
Each test consists of only an integer 18 ≤ n ≤ 10¹².

 

Output

For each test case, output r and k.

 

Sample Input

181111111

 

Sample Output

1 172 103 10

总共有N支蜡烛，要把它们放在蛋糕上 蜡烛必须放成一个同心圆的形状，圆心可以选择放一个蜡烛或者不放 从里到外第i层圆可以放k^i支蜡烛

问怎样放（选择K咯）可以使得r*k最小 如果有多种方案，选择r最小的方案 输出K和R

分析：

队友想到了用枚举+二分.....我们分析得k最小为2时，r最大也不会超过log(1e12)/log(2)=39 因此r的范围是1-39 而k的范围是2-1e12，所以应该枚举r,二分k，找出所有刚好能将蜡烛摆好的情况，记录最小值就可以了.....

代码：
二分写的太少了，实现花了好久.....

依然WA代码

#include<bits/stdc++.h>using namespace std;#define ll long longconst ll inf=(ll)1<<63-1;//不减一是负数！！！！！！！！！// 注意long long 输出格式，win测评系统用%I64d,linux 用%lldint main(){//    ll n;    while(scanf("%lld",&n)==1){        ll R,K,RK=inf;//RK初始化为(ll)1<<63爆炸        ll t,ans;        for(ll i=1;i<=64;++i){//枚举R            ll l=2,r=1e6+30;//这里的r和之前的r重名...debug n遍 ..........            ll mid;            while(l<=r){//二分K                mid=(l+r)>>1;                t=1,ans=0;                int f=0;                for(ll j=1;j<=i;++j){                    t*=mid;                    ans+=t;                    if(ans>n){f=1; break;}                }                if(f){//mid太大了                    r=mid-1;                }                else if(ans==n){                    //cout<<"i=:"<<i<<" mid=:"<<mid<<" i*mid:"<<i*mid<<" RK=:"<<RK<<endl;                    if(i*mid<RK){ R=i; K=mid; RK=i*mid;}//之前没更新RK SB.....                    else if((i*mid==RK&&i<R)){ R=i; K=mid; RK=i*mid;}//之前以为这种情况包含在上一种情况里面了...                    break; //找到了=n的接下来也有可能找到=n-1的 而这里不break二分又不能处理，所以=n和=n-1的情况分开处理比较好                }                else l=mid+1;            }        }        for(ll i=1;i<=64;++i){//枚举R            ll l=2,r=1e6+30;//这里的r和之前的r重名...debug n遍 ..........            ll mid;            while(l<=r){//二分K                mid=(l+r)>>1;                t=1,ans=0;                int f=0;                for(ll j=1;j<=i;++j){                    t*=mid;                    ans+=t;                    if(ans>n){f=1; break;}                }                if(f){//mid太大了                    r=mid-1;                }                else if(ans==(n-1)){                    //cout<<"i=:"<<i<<" mid=:"<<mid<<" i*mid:"<<i*mid<<" RK=:"<<RK<<endl;                    if(i*mid<RK){ R=i; K=mid; RK=i*mid;}                    else if((i*mid==RK&&i<R)){ R=i; K=mid; RK=i*mid;}                    break;                }                else l=mid+1;            }        }        printf("%lld %lld\n",R,K);    }return 0;}

因为参数r 以及i==1的情况没有特殊处理，疯狂WA

AC代码：

#include<bits/stdc++.h>using namespace std;#define ll long longconst ll inf=(ll)1<<63-1;//不减一是负数！！！！！！！！！// 注意long long 输出格式，win测评系统用%I64d,linux 用%lldint main(){//483MS1688K    ll n;    while(scanf("%lld",&n)==1){        ll R=1,K=n-1,RK=n-1;//RK初始化为(ll)1<<63爆炸        ll t,ans;        for(ll i=2;i<=40;++i){//枚举R            ll l=2,r=1e6+30;//这里的r和之前的r重名...debug n遍 ..........            //r=1e12+30  WA ?????????????????????????????????????            ll mid;            while(l<=r){//二分K                mid=(l+r)>>1;                t=1,ans=0;                int f=0;                for(ll j=1;j<=i;++j){                    t*=mid;                    ans+=t;                    if(ans>n){f=1; break;}                }                if(f){//mid太大了                    r=mid-1;                }                else if(ans==n){                    //cout<<"i=:"<<i<<" mid=:"<<mid<<" i*mid:"<<i*mid<<" RK=:"<<RK<<endl;                    if(i*mid<RK){ R=i; K=mid; RK=i*mid;}//之前没更新RK SB.....                    else if((i*mid==RK&&i<R)){ R=i; K=mid; RK=i*mid;}//之前以为这种情况包含在上一种情况里面了...                    break; //找到了=n的接下来也有可能找到=n-1的 而这里不break二分又不能处理，所以=n和=n-1的情况分开处理比较好                }                else l=mid+1;            }        }        for(ll i=2;i<=40;++i){//枚举R            ll l=2,r=1e6+30;//这里的r和之前的r重名...debug n遍 ..........            ll mid;            while(l<=r){//二分K                mid=(l+r)>>1;                t=1,ans=0;                int f=0;                for(ll j=1;j<=i;++j){                    t*=mid;                    ans+=t;                    if(ans>n){f=1; break;}                }                if(f){//mid太大了                    r=mid-1;                }                else if(ans==(n-1)){                    //cout<<"i=:"<<i<<" mid=:"<<mid<<" i*mid:"<<i*mid<<" RK=:"<<RK<<endl;                    if(i*mid<RK){ R=i; K=mid; RK=i*mid;}                    else if((i*mid==RK&&i<R)){ R=i; K=mid; RK=i*mid;}                    break;                }                else l=mid+1;            }        }        cout<<R<<" "<<K<<endl;//两种方式都可以        //printf("%lld %lld\n",R,K);    }return 0;}

队友1A代码，先贴过来：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <string>#include <vector>#include <stack>#include <bitset>#include <set>#include <list>#include <deque>#include <map>#include <queue>#define Mod 1000000007using namespace std;typedef long long ll;const double PI=acos(-1.0);const double eps=1e-6;const int INF=1000000000;const int maxn=100;ll n;ll bi_search(int r, ll ans){ll left=2;ll right=ans;while(left<=right){ll mid=(left+right)>>1;double t=(pow(mid*1.0,r+1)-mid)/(mid+1);if(t>((double)ans)){right=mid-1;continue;}ll temp=0;ll tt=1;for(int i=0;i<r;i++){tt*=mid;temp+=tt;}if(temp==ans)return mid;else if(temp<ans)left=mid+1;else right=mid-1;}return 0;}int main(){ll n;ll ans,k;int ar;while(scanf("%lld",&n)!=EOF){ans=n-1;ar=1;k=n-1;for(int r=2;r<=40;r++){ll t=bi_search(r,n);if(t==0)continue;if(r*t<ans){ans=r*t;ar=r;k=t;}else if(r*t==ans&&r<ar){ar=r;k=t;}}for(int r=2;r<=40;r++){ll t=bi_search(r,n-1);if(t==0)continue;if(r*t<ans){ans=r*t;ar=r;k=t;}else if(r*t==ans&&r<ar){ar=r;k=t;}}printf("%d %lld\n",ar,k);}return 0;}