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

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;

}

巴特西

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

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