Educational Codeforces Round 89 (Rated for Div. 2) A. Shovels and Swords (贪心)

标签：swa   const   ios   and   sync   set   ems   iostream   inline   

• 题意:你有\(a\)个树枝和\(b\)个钻石,\(2\)个树枝和\(1\)个钻石能造一个铁铲,\(1\)个树枝和\(2\)个钻石能造一把剑,问最多能造多少铲子和剑.

• 题解:如果\(a\le b\),若\(b\ge 2a\),那么一直取\(b\)即可,否则就要两两轮流减,即\((a+b)/3\),取个min即可.

• 代码:

  #include <iostream>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  
  int t;
  int a,b;
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);
  	cin>>t;
  	 while(t--){
  	 	cin>>a>>b;
  	 	if(a>b) swap(a,b);
  	 	int ans=min(a,(a+b)/3);
  	 	printf("%d\n",ans);
  	 }
  
      return 0;
  }
  

Educational Codeforces Round 89 (Rated for Div. 2) A. Shovels and Swords (贪心)

标签：swa   const   ios   and   sync   set   ems   iostream   inline   

原文地址：https://www.cnblogs.com/lr599909928/p/13107738.html