Team Composition: Programmers and Mathematicians 贪心（800）

题意：

给p的个数和m的个数，要求4个一组，且每组中p和m的个数≥1\geq1≥1，求最多分多少组

思路：

令p≤mp \leq mp≤m，设差值d=m−pd=m-pd=m−p，先尽量把差值抵消，每次p−1，m−3，d−2p-1，m-3，d-2p−1，m−3，d−2，组成min(d/2,min(p,m/3))min(d/2,min(p,m/3))min(d/2,min(p,m/3))组
然后差值小于2，每次就可以两个数字都分别取两个，贪心地保证a和b都尽量取完

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <unordered_set>
#include <math.h>
using namespace std;typedef long long ll;
typedef pair<int, int> PII;#define endl '\n'
#define fi first
#define se second
#define push_back
#define rep(i, l, r) for (ll i = l; i <= r; i ++ )void solve()
{int p, m; cin >> p >> m;if (p > m) swap(p, m);int d = m - p;int ans = min(d / 2, min(p, m / 3));p -= ans, m -= 3 * ans;ans += min(p / 2, m / 2);cout << ans << endl;
}int main()
{cin.tie(nullptr) -> sync_with_stdio(false);int _;cin >> _;while (_ -- )solve();return 0;
}

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <unordered_set>
#include <math.h>
using namespace std;typedef long long ll;
typedef pair<int, int> PII;#define endl '\n'
#define fi first
#define se second
#define push_back
#define rep(i, l, r) for (ll i = l; i <= r; i ++ )void solve()
{int p, m; cin >> p >> m;if (p > m) swap(p, m);int d = m - p;if (p < d / 2) cout << p << endl;elsecout << d / 2 + (p - d / 2) / 2 << endl;
}int main()
{cin.tie(nullptr) -> sync_with_stdio(false);int _;cin >> _;while (_ -- )solve();return 0;
}

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