HDOJ 4002 Find the maximum（数论+打表）

 Find the maximum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)

Total Submission(s): 2081    Accepted Submission(s): 875

Problem Description

Euler's Totient function, φ (n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n . For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
HG is the master of X Y. One day HG wants to teachers XY something about Euler's Totient function by a mathematic game. That is HG gives a positive integer N and XY tells his master the value of 2<=n<=N for which φ(n) is a maximum. Soon HG finds that this seems a little easy for XY who is a primer of Lupus, because XY gives the right answer very fast by a small program. So HG makes some changes. For this time XY will tells him the value of 2<=n<=N for which n/φ(n) is a maximum. This time XY meets some difficult because he has no enough knowledge to solve this problem. Now he needs your help.

 

Input

There are T test cases (1<=T<=50000). For each test case, standard input contains a line with 2 ≤ n ≤ 10^100.

 

Output

For each test case there should be single line of output answering the question posed above.

 

Sample Input

210100

 

Sample Output

630
Hint
If the maximum is achieved more than once, we might pick the smallest such n.  
 

 

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest 

思路：

给出一个整数n，求一个数x，x在1到n之间，并且x/φ(x)最大(其中φ(x)为x的欧拉函数)。

由欧拉函数为积性函数，即：如果

则有：

且：

则有：

要使f(x)最大，须使x含尽量多的不同素数因子。

因为数的范围给的很大，所以需要用高精度来做；并且因为数据量很大，所以一般都要通过打表来找到规律。

代码：

打表代码：

<span style="font-weight: normal;">#include <stdio.h>  #include <stdlib.h>  #include <string.h>  #include <cmath>    const int maxn = 1000;  int vis[100000];  int prime[maxn];  int ans[maxn][maxn];  void prime_table(int n)  {      int m = sqrt(n+0.5);      memset(vis,0,sizeof(vis));      for(int i = 2;i <= m;i++)      {          if(!vis[i])          {              for(int j = i*i;j <= n;j += i)              {                  vis[j] = 1;              }          }      }      for(int i = 1,j = 0;i <= n;i++)      {          if(vis[i] == 0)          {              prime[j++] = i;          }      }  }  //大整数乘法  void multiply(int a[],int n)  {      int vis = 0;      for(int i = 0;i <= maxn;i++)      {          vis += a[i] * n;          a[i] = vis % 10;          vis /= 10;      }  }    void table(int n)  {      ans[0][0] = 2;      prime_table(500);      for(int i = 1;i <= n;i++)      {          memcpy(ans[i],ans[i-1],maxn*sizeof(int));          multiply(ans[i],prime[i+1]);      }  }    int main ()  {      freopen("in.txt","r",stdin);      table(100);      for(int i = 0;i <= 52;i++)      {          int j;          for(j = 100;j >= 0;j--)              if(ans[i][j] != 0)break;          if(j >= 0)              printf("\"");          for(;j >= 0;j--)              printf("%d",ans[i][j]);          printf("\",\n");      }      return 0;  }  </span>

题AC代码：

#include <cstdio>  #include <cmath>  #include <cstdlib>  #include <cstring>  #include <iostream>  #include <algorithm>  using namespace std;    char tb[][150] = {  "2",  "6",  "30",  "210",  "2310",  "30030",  "510510",  "9699690",  "223092870",  "6469693230",  "200560490130",  "7420738134810",  "304250263527210",  "13082761331670030",  "614889782588491410",  "32589158477190044730",  "1922760350154212639070",  "117288381359406970983270",  "7858321551080267055879090",  "557940830126698960967415390",  "40729680599249024150621323470",  "3217644767340672907899084554130",  "267064515689275851355624017992790",  "23768741896345550770650537601358310",  "2305567963945518424753102147331756070",  "232862364358497360900063316880507363070",  "23984823528925228172706521638692258396210",  "2566376117594999414479597815340071648394470",  "279734996817854936178276161872067809674997230",  "31610054640417607788145206291543662493274686990",  "4014476939333036189094441199026045136645885247730",  "525896479052627740771371797072411912900610967452630",  "72047817630210000485677936198920432067383702541010310",  "10014646650599190067509233131649940057366334653200433090",  "1492182350939279320058875736615841068547583863326864530410",  "225319534991831177328890236228992001350685163362356544091910",  "35375166993717494840635767087951744212057570647889977422429870",  "5766152219975951659023630035336134306565384015606066319856068810",  "962947420735983927056946215901134429196419130606213075415963491270",  "166589903787325219380851695350896256250980509594874862046961683989710",  "29819592777931214269172453467810429868925511217482600306406141434158090",  "5397346292805549782720214077673687806275517530364350655459511599582614290",  "1030893141925860008499560888835674370998623848299590975192766715520279329390",  "198962376391690981640415251545285153602734402721821058212203976095413910572270",  "39195588149163123383161804554421175259738677336198748467804183290796540382737190",  "7799922041683461553249199106329813876687996789903550945093032474868511536164700810",  "1645783550795210387735581011435590727981167322669649249414629852197255934130751870910",  "367009731827331916465034565550136732339800312955331782619462457039988073311157667212930",  "83311209124804345037562846379881038241134671040860314654617977748077292641632790457335110",  "19078266889580195013601891820992757757219839668357012055907516904309700014933909014729740190",  "4445236185272185438169240794291312557432222642727183809026451438704160103479600800432029464270",  "1062411448280052319722448549835623701226301211611796930357321893850294264731624591303255041960530",  "256041159035492609053110100510385311995538591998443060216114576417920917800321526504084465112487730",  "64266330917908644872330635228106713310880186591609208114244758680898150367880703152525200743234420230"  };    int main()  {    //freopen("in.txt","r",stdin);      char str[150];      int t;      cin>>t;      while(t--){          scanf("%s",str);           int i=0;int str1= strlen(str);         while ( str1 > strlen(tb[ i ]) )              ++ i;          if ( strlen(tb[ i ]) == str1 && strcmp( tb[ i ], str ) > 0 )              -- i;          if ( strlen(tb[ i ]) > str1 )              -- i;          if ( str1 == 1 && strcmp( str, "6" ) >= 0 )              printf("6\n");          else if ( str1 == 1 && strcmp( str, "2" ) >= 0 )              printf("2\n");          else              printf("%s\n",tb[ i ]);      }      return 0;  }