HDU 1398 Square Coins

好吧。。 这题是母函数的同一类型题。。。。。暴力也可以过

#include<stdio.h>
int sq[18],m1[305],m2[305];
void chart(  )
{
     for( int i = 0; i < 305; ++i )
          m1[i] = m2[i] = 0;
     for( int i = 0; i < 18; ++i )
          sq[i] = i * i;
     m1[ 0 ] = 1;
     for( int i = 1; i <= 17; ++i )
     {
          for( int j = 0; j <= 300; ++j )
               for( int k = 0; ( j + k * sq[i] <= 300 ); ++k )
                    m2[j + k *sq[i]] +=  m1[j];
          for( int j = 0; j <= 300; ++j )
               m1[j] = m2[j],m2[j] = 0;
      }
 }
int main( )
{
    chart( );
    int n;
    while( scanf( "%d",&n ),n )
           printf( "%d\n",m1[n] );
    return 0;
}

下面是背包解法，没有数量限制的背包都是用完全背包解法

#include<stdio.h>
#include<string.h>
int v[18],dp[305],n;
void DP(  )
{
     memset( dp,0,sizeof( dp ) );
     for( int i = 0; i < 18; ++i )
          v[i] = i * i;
     dp[0] = 1;
     for( int i = 1; i <= 17; ++i )
          for( int j = v[i]; j <= 300; ++j )
               dp[j] += dp[j-v[i]];
 }
int main( )
{
    DP(  );
    while( scanf( "%d",&n ),n )
           printf( "%d\n",dp[n] );
    return 0;
}