HDU4002

Find the maximum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1951    Accepted Submission(s): 817

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

   
   
   
   

    
    
    
    2
10
100
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    6
30


    
    
    
    
 
     
 
      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

 

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/*
根据欧拉函数的公式，φ(x)=x(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…..(1-1/pn)

则：x/φ(x)=p1/(p1-1)*p2/(p2-1)*......*pn/(pn-1)

可以看出项越多x/φ(x)越大，且因子pi越小x/φ(x)越大，那么只需要2*3*5*7....
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//在table工程中打出这个表
char prime[ 53 ][ 101 ] = {
"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"};

int len[53];

char data[101];

int main()
{
    for ( int i = 0 ; i < 53 ; ++ i )
        len[ i ] = strlen( prime[ i ] );
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",data);
        int i = 0,l = strlen( data );
        while ( l > len[ i ] )
            ++ i;
        if ( len[ i ] == l && strcmp( prime[ i ], data ) > 0 )
            -- i;
        if ( len[ i ] > l )
            -- i;
        if ( l == 1 && strcmp( data, "6" ) >= 0 )
            printf("6\n");
        else if ( l == 1 && strcmp( data, "2" ) >= 0 )
            printf("2\n");
        else
            printf("%s\n",prime[ i ]);
    }
    return 0;
}

打表的代码

#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];

//得到素数数列prime

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("output.txt","w",stdout);
    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;
}