USACO－Subset Sums

最新推荐文章于 2024-03-19 19:37:15 发布

superxtong

最新推荐文章于 2024-03-19 19:37:15 发布

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分类专栏： dp 文章标签： dp

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本文介绍了一种使用动态规划方法解决特定等差数列子集划分问题的算法实现。针对从1到N的连续整数集合，算法能够计算出所有可能将其分为两个和相等子集的方法数量。

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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

题意：给你从1－n的连续数字，求有多少种将它们平分成两部分的方法。

等差数列求和公式：S＝(n+1)*n/2。
要平分，每一部分为S1=(n+1)*n/4,于是在开头判断S1是否为整数，不能则直接不能。
接着dp，最后要除以2，因为会算重复一次。

f[i,j]=f[i-1,j]+f[i-1,j-i] ,j-i>=0
f[i,j]=f[i-1,j] ,j-i<0

代码：

/*
ID:iam666
PROG:subset
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
using namespace std;
long long dp[800];
int n;
int main (void)
{
    //freopen("subset.in","r",stdin);
    //freopen("subset.out","w",stdout);
    cin>>n;
    if(n*(n+1)%4)
        {
            printf("0\n");
            return 0;
        }
    int re=n*(n+1)/4;
    memset(dp,0,sizeof(dp));
    for(int i=0;i<=n;i++)
    dp[0]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=re;j>=i;j--)
        {
            dp[j]=dp[j]+dp[j-i];//j比i多出来的数的分法加上原来的分法
        //  printf("%lld___\n",dp[j]);
        }
    //  printf("*******\n");
    }
    printf("%lld\n",dp[re]/2);
    return 0;

}