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
太天真了，前幾天dfs做多了，以爲又是一個二值搜索，簡單的寫了一個dfs到第四個test就開始超時了。。。哦，果然這種集合裏挑幾個數的問題應該歸爲揹包問題。設b[i][j] 是前i個數和爲j的種數。轉移方程附圖。
代碼如下：
/*
ID: gjj50201
LANG: C++
TASK: subset
*/
#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int n,sum;
int b[40][850] = {0};
int main(){
freopen("subset.in","r",stdin);
freopen("subset.out","w",stdout);
cin>>n;
if((1+n)*n/2 % 2 == 1){
cout<<"0"<<endl;
return 0;
}
b[1][1] = 1;
for(int i=2;i<=n;i++){
sum = (1+i)*i/2;
for(int j=1;j<=sum;j++){
if(j>=i)
b[i][j] = b[i-1][j-i] + b[i-1][j];
else
b[i][j] = b[i-1][j];
}
}
cout<<b[n][(1+n)*n/4]<<endl;
return 0;
}