Description

Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top.

FJ has?N?cows (1 ≤?N?≤ 20) each with some height of?H_i?(1 ≤?H_i?≤ 1,000,000 - these are very tall cows). The bookshelf has a height of?B?(1 ≤?B?≤?S, where?S?is the sum of the heights of all cows).

To reach the top of the bookshelf, one or more of the cows can stand on top of each other in a stack, so that their total height is the sum of each of their individual heights. This total height must be no less than the height of the bookshelf in order for the cows to reach the top.

Since a taller stack of cows than necessary can be dangerous, your job is to find the set of cows that produces a stack of the smallest height possible such that the stack can reach the bookshelf. Your program should print the minimal 'excess' height between the optimal stack of cows and the bookshelf.

Input

* Line 1: Two space-separated integers:?N?and?B
* Lines 2..N+1: Line?i+1 contains a single integer:?H_i

Output

* Line 1: A single integer representing the (non-negative) difference between the total height of the optimal set of cows and the height of the shelf.

Sample Input

5 16 3 1 3 5 6

Sample Output

1
題目大意 ： 有N頭牛一個書架，書架太高要站在牛身上夠書架，已知N頭牛的高度和書架的高度，一頭牛可以站在另一頭牛身上，總高度是他們的高度之和，高度和需要不小于（大于等于均可）書架高度，問符合要求的最低高度是多少，輸出該高度與書架高度的差值。 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define N 1005000 #define oo 0x3f3f3f int a[N]; int dp[N]; int vis[N]; int n,p; int main()///dp儲存牛的高度和，背包容量 v 為所有牛的高度總和，價值和體積就是每頭牛的高度。 {int t;int n,b;while(scanf("%d%d",&n,&b)!=EOF){int sum = 0;memset(a,0,sizeof(a));for(int i=1;i<=n;i++){scanf("%d",&a[i]);sum += a[i];}int Min = oo;for(int i=1;i<=n;i++){for(int j=sum;j>=a[i];j--){dp[j] = max(dp[j],dp[j-a[i]]+a[i]);if(dp[j]>=b)///是大于等于{Min = min(Min,dp[j]);}}}printf("%d\n",Min-b);}return 0; }

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轉載于:https://www.cnblogs.com/biu-biu-biu-/p/5741313.html

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