SGU 415. Necessary Coins ( 背包dp )

题意大概是:给出N个硬币, 面值为a_i, 问要凑成X元哪些硬币是不可或缺的.1 ≤ N ≤ 200, 1 ≤ x ≤ 10^4

直接枚举, 然后就是01背包了. 为了不让复杂度多乘个N, 我们就从左往右, 从右往左分别dp一次.这样判断一个硬币就是O(X).总时间复杂度O(NX)

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#include<bits/stdc++.h>

 

using namespace std;

 

const int maxn = 209;

const int maxv = 10009;

 

bool L[maxn][maxv], R[maxn][maxv];

int v[maxn], N, V;

vector<int> ans;

 

int main() {

memset(L, false, sizeof L);

memset(R, false, sizeof R);

cin >> N >> V;

for(int i = 1; i <= N; i++)

   scanf("%d", v + i);

L[0][0] = R[N + 1][0] = true;

for(int i = 1; i <= N; i++) {

for(int j = 0; j <= V; j++)

   L[i][j] = L[i - 1][j];

for(int j = V; j >= v[i]; j--)

   L[i][j] |= L[i - 1][j - v[i]];

}

for(int i = N; i; i--) {

for(int j = 0; j <= V; j++)

   R[i][j] = R[i + 1][j]; 

for(int j = V; j >= v[i]; j--)

   R[i][j] |= R[i + 1][j - v[i]];

}

for(int i = 1; i <= N; i++) {

bool F = false;

for(int j = 0; j <= V; j++)

   if(L[i - 1][j] && R[i + 1][V - j]) {

   F = true;

   break;

}

if(!F) ans.push_back(v[i]);

}

printf("%d\n", ans.size());

for(int i = 0; i < ans.size(); i++)

   printf("%d ", ans[i]);

return 0;

}

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415. Necessary Coins

Time limit per test: 1.25 second(s)
Memory limit: 262144 kilobytes

input: standard
output: standard

Vasya has been on vacation on Mars. He's a big fan of foreign coins, and thus has collected exactly one martian coin of each denomination, for a total of n coins: a₁ martian dollars, a₂ martian dollars, etc, a_n martian dollars. Unfortunately, he couldn't stand ordering the Pan Galactic Gargle Blaster at the Starport, and has to pay for it — it costs x martian dollars. Vasya is wondering which of his coins are absolutely necessary to do so (i.e., he is forced to abandon them). They don't offer change at the Starport Mars.

Input

The input file contains two integer numbers n and x (1 ≤ n ≤ 200, 1 ≤ x ≤ 10⁴), followed by n distinct integer numbersa_i (1 ≤ a_i ≤ x).

Output

On the first line of output, print the amount of denominations of coins that appear in any subset that sums to xmartian dollars. On the second line of output, print the denominations themselves, in any order, separated with single spaces. It is guaranteed that there exists at least one way to pay x martian dollars with the given coins.

Example(s)

sample input	sample output
5 18 1 2 3 5 10	2 5 10

 

转载于:https://www.cnblogs.com/JSZX11556/p/4719198.html