USACO 2.2 Subset Sums

For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical.

For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical:

• {3} and {1,2}

This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions).

If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum:

• {1,6,7} and {2,3,4,5}
• {2,5,7} and {1,3,4,6}
• {3,4,7} and {1,2,5,6}
• {1,2,4,7} and {3,5,6}

Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways.

Your program must calculate the answer, not look it up from a table.

PROGRAM NAME: subset

INPUT FORMAT

The input file contains a single line with a single integer representing N, as above.

SAMPLE INPUT (file subset.in)

7

OUTPUT FORMAT

The output file contains a single line with a single integer that tells how many same-sum partitions can be made from the set {1, 2, ..., N}. The output file should contain 0 if there are no ways to make a same-sum partition.

SAMPLE OUTPUT (file subset.out)

4

 

类型：DP

题意：将1~n，n个数分成两组，使其和相等，求一共有多少种分法。改变数字的顺序不会增加分法。

思路：很裸的背包。如果M=n*(n+1)/2是奇数，则没有分法。如果是偶数，背包容量为M/2，dp[k] += dp[k-i],(i=1,2,...,n)计算k的时候为避免重算，

需倒着进行。

代码

   
     

    
      
    
      /*
    
      
ID: superbi1
TASK: subset
LANG: C

    
      */
    
      
#include 
    
      <
    
      stdio.h
    
      >
    
      
#include 
    
      <
    
      string
    
      .h
    
      >
    
      


    
      int
    
       main()
{
 unsigned 
    
      int
    
       dp[
    
      391
    
      ];
 
    
      int
    
       i, j, n, M;
 FILE 
    
      *
    
      in
    
       
    
      =
    
       fopen(
    
      "
    
      subset.in
    
      "
    
      , 
    
      "
    
      r
    
      "
    
      );
 FILE 
    
      *
    
      out
    
       
    
      =
    
       fopen(
    
      "
    
      subset.out
    
      "
    
      , 
    
      "
    
      w
    
      "
    
      );

 fscanf(
    
      in
    
      , 
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      n);
 M 
    
      =
    
       n
    
      *
    
      (n
    
      +
    
      1
    
      )
    
      /
    
      2
    
      ;
 
    
      if
    
       (M
    
      &
    
      1
    
      ) {
 fprintf(
    
      out
    
      , 
    
      "
    
      0\n
    
      "
    
      );
 
    
      return
    
       
    
      0
    
      ;
 } 
 M 
    
      /=
    
       
    
      2
    
      ;
 memset(dp, 
    
      0
    
      , 
    
      sizeof
    
      (dp));
 dp[
    
      0
    
      ] 
    
      =
    
       
    
      1
    
      ;
 
    
      for
    
       (i
    
      =
    
      1
    
      ; i
    
      <=
    
      n; i
    
      ++
    
      ) { 
 
    
      for
    
       (j
    
      =
    
      M
    
      -
    
      i; j
    
      >=
    
      0
    
      ; j
    
      --
    
      )
 
    
      if
    
       (dp[j]) {
 dp[j
    
      +
    
      i] 
    
      +=
    
       dp[j];
 }
 }
 fprintf(
    
      out
    
      , 
    
      "
    
      %u\n
    
      "
    
      , dp[M]
    
      /
    
      2
    
      );
 
    
      return
    
       
    
      0
    
      ;
}