BestCoder Round #84 <LIS 2进制 思维>

Aaronson

 

 Accepts: 607

 

 Submissions: 1869

 Time Limit: 4000/2000 MS (Java/Others)

 

 Memory Limit: 131072/131072 K (Java/Others)

Problem Description

Recently, Peter saw the equation x_{0}+2x_{1}+4x_{2}+...+2^{m}x_{m}=nx​0​​+2x​1​​+4x​2​​+...+2​m​​x​m​​=n. He wants to find a solution (x_0,x_1,x_2,...,x_m)(x​0​​,x​1​​,x​2​​,...,x​m​​) in such a manner that \displaystyle\sum_{i=0}^{m} x_i​i=0​∑​m​​x​i​​ is minimum and every x_ix​i​​ ($0\le i\le m$) is non-negative.

Input

There are multiple test cases. The first line of input contains an integer $T$ (1 \le T \le 10^5)(1≤T≤10​5​​), indicating the number of test cases. For each test case:

The first contains two integers $n$ and $m$ (0 \le n,m \le 10^9)(0≤n,m≤10​9​​).

Output

For each test case, output the minimum value of \displaystyle\sum_{i=0}^{m} x_i​i=0​∑​m​​x​i​​.

Sample Input

101 23 25 210 210 310 413 520 411 1112 3

Sample Output

1223223232

Aaronson

 

 Accepts: 607

 

 Submissions: 1869

 Time Limit: 4000/2000 MS (Java/Others)

 

 Memory Limit: 131072/131072 K (Java/Others)

问题描述

给出一个不定方程x_{0}+2x_{1}+4x_{2}+...+2^{m}x_{m}=nx​0​​+2x​1​​+4x​2​​+...+2​m​​x​m​​=n, 找出一组解(x_0,x_1,x_2,...,x_m)(x​0​​,x​1​​,x​2​​,...,x​m​​), 使得\displaystyle\sum_{i=0}^{m} x_i​i=0​∑​m​​x​i​​最小, 并且每个x_ix​i​​ ($0\le i\le m$)都是非负的.

输入描述

输入包含多组数据, 第一行包含一个整数$T$ (1 \le T \le 10^5)(1≤T≤10​5​​)表示测试数据组数. 对于每组数据:第一行包含两个整数$n$和$m$ (0 \le n,m \le 10^9)(0≤n,m≤10​9​​).

输出描述

对于每组数据, 输出\displaystyle\sum_{i=0}^{m} x_i​i=0​∑​m​​x​i​​的最小值.

输入样例

101 23 25 210 210 310 413 520 411 1112 3

输出样例

1223223232

思路：

尽量取大的---

代码：

#include<cstdio>int shu[33];void ppp(){    shu[0]=1;    for (int i=1;i<32;i++)    shu[i]=shu[i-1]*2;}int main(){    ppp();    int t;scanf("%d",&t);    while (t--)    {        int n,m;        scanf("%d%d",&n,&m);        if (m>31) m=31;        int s=0;        for (int i=m;i>=0;i--)        {            if (n>=shu[i])            {                s+=n/shu[i];                n-=n/shu[i]*shu[i];            }        }        printf("%d\n",s);    }    return 0;}

Bellovin

 

 Accepts: 428

 

 Submissions: 1685

 Time Limit: 6000/3000 MS (Java/Others)

 

 Memory Limit: 131072/131072 K (Java/Others)

Problem Description

Peter has a sequence a_1,a_2,...,a_na​1​​,a​2​​,...,a​n​​ and he define a function on the sequence -- F(a_1,a_2,...,a_n)=(f_1,f_2,...,f_n)F(a​1​​,a​2​​,...,a​n​​)=(f​1​​,f​2​​,...,f​n​​), where f_if​i​​ is the length of the longest increasing subsequence ending with a_ia​i​​.

Peter would like to find another sequence b_1,b_2,...,b_nb​1​​,b​2​​,...,b​n​​ in such a manner that F(a_1,a_2,...,a_n)F(a​1​​,a​2​​,...,a​n​​) equals to F(b_1,b_2,...,b_n)F(b​1​​,b​2​​,...,b​n​​). Among all the possible sequences consisting of only positive integers, Peter wants the lexicographically smallest one.

The sequence a_1, a_2, ..., a_na​1​​,a​2​​,...,a​n​​ is lexicographically smaller than sequence b_1, b_2, ..., b_nb​1​​,b​2​​,...,b​n​​, if there is such number $i$ from $1$ to $n$, that a_k = b_ka​k​​=b​k​​ for $1\le k<i$ and a_i < b_ia​i​​<b​i​​.

Input

There are multiple test cases. The first line of input contains an integer $T$, indicating the number of test cases. For each test case:

The first contains an integer $n$ $(1\le n\le 100000)$ -- the length of the sequence. The second line contains $n$ integers a_1,a_2,...,a_na​1​​,a​2​​,...,a​n​​ (1 \le a_i \le 10^9)(1≤a​i​​≤10​9​​).

Output

For each test case, output $n$ integers b_1,b_2,...,b_nb​1​​,b​2​​,...,b​n​​ (1 \le b_i \le 10^9)(1≤b​i​​≤10​9​​) denoting the lexicographically smallest sequence.

Sample Input

311055 4 3 2 131 3 5

Sample Output

Copy

11 1 1 1 11 2 3

Bellovin

 

 Accepts: 428

 

 Submissions: 1685

 Time Limit: 6000/3000 MS (Java/Others)

 

 Memory Limit: 131072/131072 K (Java/Others)

问题描述

Peter有一个序列a_1,a_2,...,a_na​1​​,a​2​​,...,a​n​​. 定义F(a_1,a_2,...,a_n)=(f_1,f_2,...,f_n)F(a​1​​,a​2​​,...,a​n​​)=(f​1​​,f​2​​,...,f​n​​), 其中f_if​i​​是以a_ia​i​​结尾的最长上升子序列的长度.Peter想要找到另一个序列b_1,b_2,...,b_nb​1​​,b​2​​,...,b​n​​使得F(a_1,a_2,...,a_n)F(a​1​​,a​2​​,...,a​n​​)和F(b_1,b_2,...,b_n)F(b​1​​,b​2​​,...,b​n​​)相同. 对于所有可行的正整数序列, Peter想要那个字典序最小的序列.序列a_1, a_2, ..., a_na​1​​,a​2​​,...,a​n​​比b_1, b_2, ..., b_nb​1​​,b​2​​,...,b​n​​字典序小, 当且仅当存在一个正整数$i$ $(1\le i\le n)$满足对于所有的$k$ $(1\le k<i)$都有a_k = b_ka​k​​=b​k​​并且a_i < b_ia​i​​<b​i​​.

输入描述

输入包含多组数据, 第一行包含一个整数$T$表示测试数据组数. 对于每组数据:第一行包含一个整数$n$ $(1\le n\le 100000)$表示序列的长度. 第二行包含$n$个整数a_1,a_2,...,a_na​1​​,a​2​​,...,a​n​​ (1 \le a_i \le 10^9)(1≤a​i​​≤10​9​​).

输出描述

对于每组数据, 输出$n$个整数b_1,b_2,...,b_nb​1​​,b​2​​,...,b​n​​ (1 \le b_i \le 10^9)(1≤b​i​​≤10​9​​)表示那个字典序最小的序列.

输入样例

311055 4 3 2 131 3 5

输出样例

11 1 1 1 11 2 3

运用LIS---求法，求出以每一位结尾的最长链长度---

每个的长度的排序是最优小字典序--

代码：

#include<cstdio>#include<cstring>#define MA 100100int t,n,ge;int kp[MA];int shu[MA];int chu[MA];int zhao(int xx){    int l=0,r=ge-1,mid,ans=0;    while (r>=l)    {        mid=(l+r)/2;        if (kp[mid]>=xx)        {            ans=mid;            r=mid-1;        }        else        l=mid+1;    }    return ans;}int main(){    scanf("%d",&t);    while (t--)    {        scanf("%d",&n);ge=0;        for (int i=0;i<n;i++)        {            scanf("%d",&shu[i]);            if (ge==0)            {                chu[i]=1;                kp[ge++]=shu[i];            }            else            {                if (shu[i]>kp[ge-1])                {                    chu[i]=ge+1;                    kp[ge]=shu[i];                    ge++;                }                else                {                    chu[i]=zhao(shu[i]);                  //  if (chu[i]!=-1)                   // {                        kp[chu[i]]=shu[i];                        chu[i]++;                   /* }                    else                    {                        chu[i]=1;                    }*/                }            }        }        for (int i=0;i<n-1;i++)        printf("%d ",chu[i]);        printf("%d\n",chu[n-1]);    }    return 0;}

Dertouzos

 

 Accepts: 76

 

 Submissions: 1357

 Time Limit: 7000/3500 MS (Java/Others)

 

 Memory Limit: 131072/131072 K (Java/Others)

Problem Description

A positive proper divisor is a positive divisor of a number $n$, excluding $n$ itself. For example, 1, 2, and 3 are positive proper divisors of 6, but 6 itself is not.

Peter has two positive integers $n$ and $d$. He would like to know the number of integers below $n$ whose maximum positive proper divisor is $d$.

Input

There are multiple test cases. The first line of input contains an integer $T$ (1 \le T \le 10^6)(1≤T≤10​6​​), indicating the number of test cases. For each test case:

The first line contains two integers $n$ and $d$ (2 \le n, d \le 10^9)(2≤n,d≤10​9​​).

Output

For each test case, output an integer denoting the answer.

Sample Input

910 210 310 410 510 610 710 810 9100 13

Sample Output

Copy

121000004

Dertouzos

 

 Accepts: 76

 

 Submissions: 1357

 Time Limit: 7000/3500 MS (Java/Others)

 

 Memory Limit: 131072/131072 K (Java/Others)

问题描述

正整数$x$称为$n$的positive proper divisor, 当且仅当$x|n$并且$1\le x<n$. 例如, 1, 2, 和3是6的positive proper divisor, 但是6不是.Peter给你两个正整数$n$和$d$. 他想要知道有多少小于$n$的整数, 满足他们的最大positive proper divisor恰好是$d$.

输入描述

输入包含多组数据, 第一行包含一个整数$T$ (1 \le T \le 10^6)(1≤T≤10​6​​)表示测试数据组数. 对于每组数据:第一行包含两个整数$n$和$d$ (2 \le n, d \le 10^9)(2≤n,d≤10​9​​).

输出描述

对于每组数据, 输出一个整数.

输入样例

910 210 310 410 510 610 710 810 9100 13

输出样例

121000004

BC官方解析：

随便推导下, 令$y=xd$, 如果$d$是$y$的maximum positive proper divisor, 显然要求$x$是$y$的最小质因子. 令$mp(n)$表示$n$的最小质因子, 那么就有$x\le mp(d)$, 同时有$y<n$, 那么x \le \lfloor \frac{n-1}{d} \rfloorx≤⌊​d​​n−1​​⌋. 于是就是计算有多少个素数$x$满足x \le \min{mp(d), \lfloor \frac{n-1}{d} \rfloor}x≤min{mp(d),⌊​d​​n−1​​⌋}.

当$d$比较大的时候, \lfloor \frac{n-1}{d} \rfloor⌊​d​​n−1​​⌋比较小, 暴力枚举$x$即可. 当$d$比较小的时候, 可以直接预处理出答案. 阈值设置到10^6 \sim 10^710​6​​∼10​7​​都可以过.

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longint kp,shu[350000],ge[350000];void da(){    kp=0;    memset(shu,0,sizeof(shu));    memset(ge,0,sizeof(ge));    int i,j;    for (i=2;i<32000;i++)    {        if (shu[i]==0)        {            ge[kp++]=i;            for (j=i*i;j<32000;j+=i)                shu[j]=1;        }    }}int main(){    da();    int t,n,d;scanf("%d",&t);    while (t--)    {        scanf("%d%d",&n,&d);        n=(n-1)/d;        n=min(n,d);        for (int i=0;i<kp;i++)        {            if (d%ge[i]==0||ge[i]>n)//质数最多查一个d的因数---且最大到n--            {                if (ge[i]>n)                printf("%d\n",i);                else                printf("%d\n",++i);                break;            }        }    }    return 0;}