#106 (div.2) A. Business trip

標籤：codeforces   貪進法   

1.題目描述：點擊開啟連結

2.解題思路：本題利用貪進法解決。首先對數組排序，由大到小選擇，恰好大於等於k時候停止即可。如果全部相加仍然小於k，則無解。

3.代碼：

#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;#define me(s) memset(s,0,sizeof(s))typedef long long ll;typedef unsigned int uint;typedef unsigned long long ull;typedef pair <int, int> P;const int N=20;int a[N];int k;int main(){    while(~scanf("%d",&k))    {        for(int i=1;i<=12;i++)            scanf("%d",&a[i]);        sort(a+1,a+12+1);        int st=12,sum=0;        int ans=-1;        while(sum<k&&st>=1)        {            sum+=a[st--];        }        if(sum>=k&&st>=0)ans=12-st;        printf("%d\n",ans);    }}

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#106 (div.2) A. Business trip