CF# Educational Codeforces Round 3 A. USB Flash Drives
2 seconds
256 megabytes
standard input
standard output
Sean is trying to save a large file to a USB flash drive. He has n USB flash drives with capacities equal to a1, a2, ..., an megabytes. The file size is equal to m megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
The first line contains positive integer n (1 ≤ n ≤ 100) — the number of USB flash drives.
The second line contains positive integer m (1 ≤ m ≤ 105) — the size of Sean's file.
Each of the next n lines contains positive integer ai (1 ≤ ai ≤ 1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all ai is not less than m.
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
3
5
2
1
3
2
3
6
2
3
2
3
2
5
5
10
1
In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.
除法
/**
Create By yzx - stupidboy
*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
#include <iomanip>
using namespace std;
typedef long long LL;
typedef double DB;
#define MIT (2147483647)
#define INF (1000000001)
#define MLL (1000000000000000001LL)
#define sz(x) ((int) (x).size())
#define clr(x, y) memset(x, y, sizeof(x))
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define mk make_pair inline int Getint()
{
int Ret = ;
char Ch = ' ';
bool Flag = ;
while(!(Ch >= '' && Ch <= ''))
{
if(Ch == '-') Flag ^= ;
Ch = getchar();
}
while(Ch >= '' && Ch <= '')
{
Ret = Ret * + Ch - '';
Ch = getchar();
}
return Flag ? -Ret : Ret;
} const int N = ;
int n, m, arr[N]; inline void Input()
{
scanf("%d", &n);
scanf("%d", &m);
for(int i = ; i < n; i++) scanf("%d", &arr[i]);
} inline void Solve()
{
sort(arr, arr + n, greater<int>());
int ans = , cnt = ;
for(int i = ; i < n; i++)
{
if(cnt >= m) break;
ans++, cnt += arr[i];
} cout << ans << endl;
} int main()
{
freopen("a.in", "r", stdin);
Input();
Solve();
return ;
}
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