# USACO 2.2 Subset Sums集合 解题报告

USACO 2.2 Subset Sums集合
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
Description

{3} and {1,2}

{1,6,7} and {2,3,4,5} {注 1+6+7=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}

Input

Output

Sample Input
7
Sample Output
4

for(i = 2; i <= n; i++)
for(j = 1; j <= tot; j++)
…..

C语言:

/*
LANG: C
ID: zqy11001
PROG: subset
*/
#include <stdio.h>

#define min(a, b) ((a) < (b) ? (a) : (b))

int f[40][390];

int main(void)
{
int n, tot;
int i, j,t;
freopen(subset.in, r, stdin);
freopen(subset.out, w, stdout);
scanf(%d, &n);
tot = n*(n + 1) >> 1;
if(tot & 1){
printf(\\n);
return 0;
}
tot = tot >> 1;
f[1][1] = 1;
for(i = 2; i <= n; i++){
for(j = 1; j <= tot; j++){
f[i][j] = f[i - 1][j];
if(j - i >= 0){
f[i][j] += f[i - 1][j - i];
}
}
}
printf(%d\\n, f[n][tot]);
return 0;
}