母函数

晚上写了几道简单的母函数。

母函数万变不离其宗。。

1.HDU1171

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

int value[2000], num[2000];
int c1[250010], c2[250010];
int sum = 0;

void fun(int N)
{
  int cnt = num[1];
  memset(c1, 0, sizeof(c1));
  memset(c2, 0, sizeof(c2));
  for( int i = 0; cnt >= 0; i += value[1], cnt--)
  {
         c1[i] = 1;
  }
  for( int i = 2; i <= N; i++)
  {
     for( int j = 0; j <= sum; j++)
     {  
       
        for(int k = 0,  cnt = num[i]; k + j <= sum && cnt >= 0; k += value[i], cnt--)
        {
             c2[k+j] += c1[j]; 
     
        }
        
     }
     for( int k = 0; k <= sum; k++)
     {
         c1[k] = c2[k];
         c2[k] = 0;
     }
  
  }
  for( int i = sum / 2 ; i <= sum; i++)
  {
     if( c1[i] && c1[sum - i] && i >= sum - i)
     {
       printf("%d %d\n", i, sum - i);
       return;
     }
  }
}

int main( )
{
  int N;
  while( scanf("%d",&N) != EOF )
  {
    sum = 0;
    if( N < 0 )
        break;
    memset(num, 0, sizeof(num));
    memset(value, 0, sizeof(value));
    for( int i = 1; i <= N; i++)
    {
      scanf("%d%d",&value[i],&num[i]);
      sum += value[i] * num[i];
    }
    fun(N);
  }
}

2.HDU1398

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

int c1[1000], c2[1000];

void fun(  )
{
  memset(c1, 0, sizeof(c1));
  memset(c2, 0, sizeof(c2));
  for( int i = 0; i <= 300; i++)
  {
     c1[i] = 1;
  }
  for( int i = 2; i <= 17; i++)
  {
     for( int j = 0; j <= 300; j++)
     {
        for(int k = 0; k + j <= 300; k += i * i)
        {
             c2[k + j] += c1[j];    
        }
     }
    for( int k = 0; k <= 300; k++)
    {
        c1[k] = c2[k];
        c2[k] = 0;
    }
  }
}

int main( )
{
  int N;
  fun( );
  while( scanf("%d",&N) != EOF )
  {
    printf("%d\n",c1[N]);
  }
  return 0;
}

3.HDU1028

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

int c1[400], c2[400];

void fun(int N)
{
  memset(c1, 0, sizeof(c1));
  memset(c2, 0, sizeof(c2));
  for( int i = 0; i <= N; i++)
  {
    c1[i] = 1;
  }   
  for( int i = 2; i <= N; i++)
  {
     for( int j = 0; j <= N; j++)
     {
       for( int k = 0; k + j <= N; k += i)
       {
           c2[k + j] += c1[j];
       }
     }
     for( int k = 0; k <= N; k++)
     {
       c1[k] = c2[k];
       c2[k] = 0;
     }
  }
  
}

int main( )
{
  int N;
  fun(300);
  while( scanf("%d",&N) != EOF )
  {
    printf("%d\n",c1[N]);
  }
  return 0;
}

4.HDU2082

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

int c1[55000], c2[55000];
int num[10000];

void fun( )
{ 
  int sum = 0;
  memset(c1, 0, sizeof(c1));
  memset(c2, 0, sizeof(c2));
  for( int i = 0; i <= num[1]; i++)
    c1[i] = 1;
  for( int i = 2; i <= 26; i++)
  {
     for( int j = 0; j <= 50; j++)
       for( int k = 0, cnt = num[i]; k + j <= 50, cnt >= 0; k += i, cnt--)
       {
            c2[k + j] += c1[j];     
       } 
     for( int k = 0; k <= 50; k++)
     {  
        c1[k] = c2[k];
        c2[k] = 0;
     
     }
  
  }
  for( int i = 1; i <= 50; i++)
  {
      sum += c1[i];
  }
  printf("%d\n", sum);     
}

int main( )
{
  int N;
  scanf("%d", &N);
  while( N--)
  {
    for(int i = 1; i <= 26; i++)
    {  
       scanf("%d",&num[i]);
    } 
    fun( );
  }
}

5。 HDU 1085

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

int value[10] = {0, 1, 2, 5};
int num[10];
int c1[100000];
int c2[100000];
int sum = 0;

void fun( )
{
  memset(c1, 0, sizeof(c1));
  memset(c2, 0, sizeof(c2));
  for( int i = 0; i <= num[1]; i++)
  {
    c1[i] = 1;
  }
  for( int i = 2; i <= 3; i++)
  {
    for( int j = 0; j <= sum; j++)
    {
        for(int k = 0, cnt = num[i]; k + j <= sum, cnt >= 0; k += value[i],cnt-- ) 
        {
           c2[k + j] += c1[j];
        }
    }
    for( int k = 0; k <= sum; k++)
    {
        c1[k] = c2[k];
        c2[k] = 0;
    }
  
  }
  for( int i = 1; i <= sum + 10; i++)
  {
    if( !c1[i] )
    {
      printf("%d\n",i);
      return;
    }
  }
  
}
int main( )
{
  while( scanf("%d%d%d",&num[1], &num[2], &num[3]) != EOF )
  {  
     sum = 0;
     if(num[1] == 0 && num[2] == 0 && num[3] == 0)
        break;
     sum += num[1]  + num[2] * 2 + num[3] * 5;
     fun(  ); 
  }
}