USACO: Subset Sums

Subset Sums
JRM

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

 

 

思路：

题目是问，一个连续整数集合，如{ 1， 2， 3， 4， 5 }，总共可以分解成多少对和一样的子集。

先排除一种情况：如果这个数列之和为奇数，那么不可以形成平分集，直接输出0。

转化一下问题，把数列和除以二，得到目标值，即寻找在数列范围内的、和等于目标值的集合数，再除以二即得到答案。

 

如何寻找这样的集合个数？我采用一个数组来记录，每次增加一个新元素，然后更新对应的区域

一个小技巧：为了防止发生重复，应该逆序，详见代码

/* ID: geterns1 PROG: subset LANG: C++ */ #include <iostream> #include <fstream> #include <cstring> using namespace std; unsigned sub[391] = {}; //记录和等于下标的集合个数 int main(){ ifstream ifs("subset.in"); ofstream ofs("subset.out"); short n, tar; ifs >> n; short tmp; tar = (1 + n) * n / 2; if ((tar & 1) == 1) ofs << "0" << endl; else{ tar /= 2; sub[0] = 1; //假设{ 0 }也是一种组合 for (short i = 1; i <= n; i ++){ //每增加一个新元素i，i ~ (i * i + i) / 2 之间的组合方案有更新 //当然，真正需要更新的是不大于目标值的区域 tmp = (i * i + i) / 2; if (tar < tmp) tmp = tar; //逆序，保证不重复：例如i = 3时，3先从{ 1 + 2 }变为{ 1 + 2， 3 }，之后6由3的组合方案从{ }变为{ 1 + 2 + 3， 3 + 3 }，产生重复 //反过来先把6从{ }变为{ 1 + 2 + 3 }，再把3从{ 1 + 2 }变为{ 1 + 2， 3 }即可 for (short j = tmp; j >= i; j --) sub[j] += sub[j - i]; } ofs << sub[tar] / 2 << endl; } ifs.close(); ofs.close(); return 0; }

 

USER: Geterns Liu [geterns1]
TASK: subset
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3012 KB]
   Test 2: TEST OK [0.011 secs, 3012 KB]
   Test 3: TEST OK [0.000 secs, 3012 KB]
   Test 4: TEST OK [0.022 secs, 3012 KB]
   Test 5: TEST OK [0.000 secs, 3012 KB]
   Test 6: TEST OK [0.011 secs, 3012 KB]
   Test 7: TEST OK [0.011 secs, 3012 KB]

All tests OK.
Your program ('subset') produced all correct answers!  This is your
submission #4 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 -------
7
------- test 2 -------
15
------- test 3 -------
24
------- test 4 -------
31
------- test 5 -------
36
------- test 6 -------
39
------- test 7 -------
37

Keep up the good work!

Thanks for your submission!