poj-3627

题意:

输入一个N和B N为牛的个数,B为高度,然后输入N头牛,为最少多少头牛加起来高度大于B

解题思路:

排序,贪心

具体代码:

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
int n,b;
int num[20005];
int sum=0;
cin>>n>>b;
for(int i=0;i<n;i++)
cin>>num[i];
sort(num,num+n);
int xx=n;
while(sum<b)
{
xx--;
sum+=num[xx];
}
cout<<n-xx<<endl;
// system("pause");
return 0;
}

时间: 2024-10-05 04:40:50

poj-3627的相关文章

poj 3627 Monthly Expense

题意: 我没看懂题意... 有n个数字,要把他们分成m组,每组都是连续的几个数字,要求使数字和最大的组 最小 解题思路: 二分最小数字和 判断是否能够分成至少M组 下界是max(a[1] ~a[n]) 上界是a[1] + a[2] + ....+ a[n]; code: #include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<vector>

POJ 3627 Bookshelf (贪心)

Description Farmer John recently bought a bookshelf for cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top. Each of the N cows (1 ≤ N ≤ 20,000) has some height of Hi (1 ≤ Hi ≤ 10,000) and a t

POJ - 3186 Treats for the Cows (区间DP)

题目链接:http://poj.org/problem?id=3186 题意:给定一组序列,取n次,每次可以取序列最前面的数或最后面的数,第n次出来就乘n,然后求和的最大值. 题解:用dp[i][j]表示i~j区间和的最大值,然后根据这个状态可以从删前和删后转移过来,推出状态转移方程: dp[i][j]=max(dp[i+1][j]+value[i]*k,dp[i][j-1]+value[j]*k) 1 #include <iostream> 2 #include <algorithm&

POJ 2533 - Longest Ordered Subsequence(最长上升子序列) 题解

此文为博主原创题解,转载时请通知博主,并把原文链接放在正文醒目位置. 题目链接:http://poj.org/problem?id=2533 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK)

POJ——T2271 Guardian of Decency

http://poj.org/problem?id=2771 Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 5932   Accepted: 2463 Description Frank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he is

POJ——T2446 Chessboard

http://poj.org/problem?id=2446 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18560   Accepted: 5857 Description Alice and Bob often play games on chessboard. One day, Alice draws a board with size M * N. She wants Bob to use a lot of c

poj 1088 滑雪 DP(dfs的记忆化搜索)

题目地址:http://poj.org/problem?id=1088 题目大意:给你一个m*n的矩阵 如果其中一个点高于另一个点 那么就可以从高点向下滑 直到没有可以下滑的时候 就得到一条下滑路径 求最大的下滑路径 分析:因为只能从高峰滑到低峰,无后效性,所以每个点都可以找到自己的最长下滑距离(只与自己高度有关).记忆每个点的最长下滑距离,当有另一个点的下滑路径遇到这个点的时候,直接加上这个点的最长下滑距离. dp递推式是,dp[x][y] = max(dp[x][y],dp[x+1][y]+

POJ 1385 计算几何 多边形重心

链接: http://poj.org/problem?id=1385 题意: 给你一个多边形,求它的重心 题解: 模板题,但是不知道为啥我的结果输出的确是-0.00 -0.00 所以我又写了个 if (ans.x == 0) ans.x = 0 感觉好傻逼 代码: 1 #include <map> 2 #include <set> 3 #include <cmath> 4 #include <queue> 5 #include <stack> 6

POJ 1741 Tree(树的点分治,入门题)

Tree Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 21357   Accepted: 7006 Description Give a tree with n vertices,each edge has a length(positive integer less than 1001).Define dist(u,v)=The min distance between node u and v.Give an in

poj 1655 树的重心

Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13178   Accepted: 5565 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or m