Index of super-prime
Problem's Link
Mean:
如果一个素数所在的位置还是素数,那么这个素数就是超级素数,比如3在第2位置,那么3就是超级素数.
现在给你一个数,求出来这个数由最少的超级素数的和组成,输出这个超级素数.
analyse:
很简单的完全背包,不需要二进制压缩,也不必考虑容量.
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-01-08-10.51
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long( LL);
typedef unsigned long long( ULL);
const double eps( 1e-8);
#define REP( i, n ) \
for ( int i = 0; i < (n); i++ )
#define FOR( i, b, e ) \
for ( int i = (b); i <= (e); i++ )
const int
MAXN = 10001 ,
oo = ( int) 1e9;
int N;
int cant;
int from [ MAXN ];
int best [ MAXN ];
vector < int > SP;
int prime( int x )
{
if ( x < 2 ) return false;
if ( x == 2 || x == 3 )
return true;
if ( x % 2 == 0 || x % 3 == 0 )
return false;
for ( int i = 6; ( i - 1) * ( i - 1) <= x; i += 6 )
if ( x % ( i - 1) == 0 || x % ( i + 1) == 0)
return false;
return true;
}
int main()
{
scanf( "%d" , &N );
FOR( i , 2 , N )
if ( prime( i ) )
{
cant ++;
if ( prime( cant ) )
SP . push_back( i );
}
//printf( "%d\n", SP.size() );
fill( best , best + N + 1 , oo );
best [ 0 ] = 0;
REP( i , SP . size() )
{
int val = SP [ i ];
FOR( j , val , N )
if ( best [ j - val ] + 1 < best [ j ] )
{
best [ j ] = best [ j - val ] + 1;
from [ j ] = i;
}
}
if ( best [N ] == oo )
printf( "0 \n " );
else
{
printf( "%d \n " , best [N ] );
int val = N;
while ( val > 0 )
{
printf( "%d " , SP [ from [ val ] ] );
val -= SP [ from [ val ] ];
}
printf( " \n " );
}
return 0;
}
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-01-08-10.51
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long( LL);
typedef unsigned long long( ULL);
const double eps( 1e-8);
#define REP( i, n ) \
for ( int i = 0; i < (n); i++ )
#define FOR( i, b, e ) \
for ( int i = (b); i <= (e); i++ )
const int
MAXN = 10001 ,
oo = ( int) 1e9;
int N;
int cant;
int from [ MAXN ];
int best [ MAXN ];
vector < int > SP;
int prime( int x )
{
if ( x < 2 ) return false;
if ( x == 2 || x == 3 )
return true;
if ( x % 2 == 0 || x % 3 == 0 )
return false;
for ( int i = 6; ( i - 1) * ( i - 1) <= x; i += 6 )
if ( x % ( i - 1) == 0 || x % ( i + 1) == 0)
return false;
return true;
}
int main()
{
scanf( "%d" , &N );
FOR( i , 2 , N )
if ( prime( i ) )
{
cant ++;
if ( prime( cant ) )
SP . push_back( i );
}
//printf( "%d\n", SP.size() );
fill( best , best + N + 1 , oo );
best [ 0 ] = 0;
REP( i , SP . size() )
{
int val = SP [ i ];
FOR( j , val , N )
if ( best [ j - val ] + 1 < best [ j ] )
{
best [ j ] = best [ j - val ] + 1;
from [ j ] = i;
}
}
if ( best [N ] == oo )
printf( "0 \n " );
else
{
printf( "%d \n " , best [N ] );
int val = N;
while ( val > 0 )
{
printf( "%d " , SP [ from [ val ] ] );
val -= SP [ from [ val ] ];
}
printf( " \n " );
}
return 0;
}