codefroces 939 A - C

最新推荐文章于 2024-01-22 16:24:53 发布

xmudl

最新推荐文章于 2024-01-22 16:24:53 发布

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本文链接：https://blog.youkuaiyun.com/q3818908/article/details/79625083

本文解析了三道算法竞赛题目：三角恋关系判断、选择盒子装松鼠以减少剩余数量及便利性最大化的路径选择问题。通过高效算法实现，介绍了如何在特定约束下找到最优解。

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A. Love Triangle

给你一组数判断是否存在三角恋关系.

#include <bits/stdc++.h>
//#define LOCAL_DEFINE
using namespace std;

vector<int> v(5010);
int n;

int main(void) {
    ios::sync_with_stdio(false);cin.tie(0);
#ifdef LOCAL_DEFINE
    freopen("input.txt", "rt", stdin);
#endif

    cin >> n;
    for(int i = 1; i <= n; ++i) cin >> v[i];
    for(int i = 1; i <= n; ++i) {
        int a = i;
        int b = v[a];
        int c = v[b];
        int d = v[c];
        if(a == d) cout << "YES" << endl, exit(0);
    } 
    cout << "NO" << endl;


#ifdef LOCAL_DEFINE
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
        return 0;
}

B. Hamster Farm

给你松鼠的数量和盒子的种类、及每种盒子的容量，问你选哪种盒子多少个使得剩余松鼠最少（盒子必须装满，装不满放弃）

#include <bits/stdc++.h>
//#define LOCAL_DEFINE
using namespace std;

typedef unsigned long long ull;
ull n, k, v;
ull ans1, ans2;
ull mi = (ull)1e18 + 10;

int main(void) {
    ios::sync_with_stdio(false);cin.tie(0);
#ifdef LOCAL_DEFINE
    freopen("input.txt", "rt", stdin);
#endif

    cin >> n >> k;
    for(ull i = 1; i <= k; i++) {
        cin >> v;
        ull temp = n % v;
        if(temp < mi) {
            ans1 = i;
            ans2 = n / v;
            mi = temp;
        }
    }
    cout << ans1 <<" " << ans2 << endl; 

#ifdef LOCAL_DEFINE
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
        return 0;
}

C. Convenient For Everybody

描述略
题目给的标签是binary search、two pointers. 反正我没用到二分。
用类似维护区间和的做法使复杂度达到O (n) ,注意一下边界和多解情况要求输出最小值（没注意wa两次）
-

#include <bits/stdc++.h>
//#define LOCAL_DEFINE
using namespace std;

typedef long long ll;
int n;
int s, f, d;
int ans = 0x3f3f3f3f;
const int maxn = (int)2e5 + 10;
vector<ll> v(maxn), dp(maxn);
int main(void) {
    ios::sync_with_stdio(false);cin.tie(0);
#ifdef LOCAL_DEFINE
    freopen("input.txt", "rt", stdin);
#endif

    cin >> n;
    for(int i = 1; i <= n; i++) cin >> v[i];
    for(int i = n + 1; i <= 2 * n; i++) v[i] = v[i - n];
    cin >> s >> f;
    d = f - s;
    for(int i = n + 1; i > n + 1 - d; i--) dp[1] += v[i];
    for(int i = 2; i <= n; i++) {
        dp[i] = dp[i - 1] + v[i + n] - v[i + n - d];
    }
    int id = *max_element(dp.begin() + 1, dp.begin() + 1 + n);
    for(int i = 1; i <= n; ++i) {
        if(dp[i] != id) continue;
        int st = (i + n) - d + 1;
        if(st > n) st -= n;
        if(st == 1) ans = min(ans, s);
        else {
            int dis = st - 1;
            int temp = s + n - dis;
            if(temp > n) temp -= n;
            ans = min(ans, temp);
        }
    }
    cout << ans << endl;

#ifdef LOCAL_DEFINE
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
        return 0;
}