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

Posted 2020-12-10 lr599909928

• 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(ale b),若(bge 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
  ",ans);
  	 }
  
      return 0;
  }