[Kick Start 2020] Round A 1.Allocation
1. 題目
Problem
There are N houses for sale. The i-th house costs Ai dollars to buy. You have a budget of B dollars to spend.
What is the maximum number of houses you can buy?
Input
The first line of the input gives the number of test cases, T. T test cases follow. Each test case begins with a single line containing the two integers N and B. The second line contains N integers. The i-th integer is Ai, the cost of the i-th house.
Output
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the maximum number of houses you can buy.
Limits
Time limit: 15 seconds per test set.
Memory limit: 1GB.
1 ≤ T ≤ 100.
1 ≤ B ≤ 105.
1 ≤ Ai ≤ 1000, for all i.
Test set 1
1 ≤ N ≤ 100.
Test set 2
1 ≤ N ≤ 105.
Sample
Input
Output
3
4 100
20 90 40 90
4 50
30 30 10 10
3 300
999 999 999
Case #1: 2
Case #2: 3
Case #3: 0
In Sample Case #1, you have a budget of 100 dollars. You can buy the 1st and 3rd houses for 20 + 40 = 60 dollars.
In Sample Case #2, you have a budget of 50 dollars. You can buy the 1st, 3rd and 4th houses for 30 + 10 + 10 = 50 dollars.
In Sample Case #3, you have a budget of 300 dollars. You cannot buy any houses (so the answer is 0).
Note: Unlike previous editions, in Kick Start 2020, all test sets are visible verdict test sets, meaning you receive instant feedback upon submission.
2. 思路
- 下次感覺自己能做的更好系列，畢竟這個是一個月一次的比賽，堅持
- 思路：貪心原則，買便宜的house就能買很多
- 時間複雜度nlogn
3. 代碼
#include<iostream>
#include<vector>
#include <algorithm>
using namespace std;
int main(){
int testNum;
cin >> testNum;
for (int i = 0; i < testNum; i++) {
int houseNum;
int budget;
cin >> houseNum >> budget;
vector<int> prices;
for (int j = 0; j < houseNum; j++) {
int onePrice;
cin >> onePrice;
prices.push_back(onePrice);
}
sort(prices.begin(), prices.end());
int oneres = 0;
for (int j = 0; j < houseNum; j++) {
if (budget >= prices[j]) {
oneres++;
budget -= prices[j];
} else
break;
}
cout << "Case #" << i+1 << ": " << oneres << endl;
}
return 0;
}