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。

/*
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;
}