#106 (Div.2) A. Business trip

1. Title Description: Click to open the link

2. Solving ideas: The problem is solved by greedy method. First, the order of the array, from large to small selection, exactly greater than or equal to K stop. If all additions are still less than K, then no solution.

3. Code:

#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