687C: The values you can make
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Codeforces Round #360 Editorial [+ Challenges!]
C. The Values You Can Make
Pari wants to buy an expensive chocolate from Arya. She has n coins, the value of the i-th coin isci. The price of the chocolate isk, so Pari will take a subset of her coins with sum equal tok and give it to Arya.
Looking at her coins, a question came to her mind: after giving the coins to Arya, what values does Arya can make with them? She is jealous and she doesn't want Arya to make a lot of values. So she wants to know all the valuesx, such that Arya will be able to make x using some subset of coins with the sumk.
Formally, Pari wants to know the values x such that there exists a subset of coins with the sumk such that some subset of this subset has the sumx, i.e. there is exists some way to pay for the chocolate, such that Arya will be able to make the sumx using these coins.
The first line contains two integers n andk (1 ≤ n, k ≤ 500) — the number of coins and the price of the chocolate, respectively.
Next line will contain n integers c1, c2, ..., cn (1 ≤ ci ≤ 500) — the values of Pari's coins.
It's guaranteed that one can make value k using these coins.
First line of the output must contain a single integer q— the number of suitable values x. Then printq integers in ascending order — the values that Arya can make for some subset of coins of Pari that pays for the chocolate.
6 185 6 1 10 12 2
160 1 2 3 5 6 7 8 10 11 12 13 15 16 17 18
3 5025 25 50
30 25 50
Hint
Use dynamic programming.
Solution
Let dpi, j, k be true if and only if there exists a subset of the firsti coins with sumj, that has a subset with sumk. There are 3 cases to handle:
- The i-th coin is not used in the subsets.
- The i-th coin is used in the subset to makej, but it's not used in the subset of this subset.
- The i-th coin is used in both subsets.
So dpi, j, k is equal todpi - 1, ji, kOR dpi - 1, j - ci, kOR dpi - 1, j - ci, k - ci.
The complexity is O(nk2).
C++ code
// - -- --- ---- -----be name khoda----- ---- --- -- -\\#include <bits/stdc++.h>using namespace std;inline int in() { int x; scanf("%d", &x); return x; }const int N = 505;bool dp[2][N][N];int main(){int n, k;cin >> n >> k;dp[0][0][0] = 1;for(int i = 1; i <= n; i++){int now = i % 2;int last = 1 - now;int x = in();for(int j = 0; j <= k; j++)for(int y = 0; y <= j; y++){dp[now][j][y] = dp[last][j][y];if(j >= x){ dp[now][j][y] |= dp[last][j - x][y]; if(y >= x) dp[now][j][y] |= dp[last][j - x][y - x];}}}vector <int> res;for(int i = 0; i <= k; i++)if(dp[n % 2][k][i])res.push_back(i);cout << res.size() << endl;for(int x : res)cout << x << " ";cout << endl;}
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