解题思路

由于三种国家都有获胜的可能，所以我们需要分别枚举 $X,Y,Z$ 获胜的情况。

设 $X$ 获胜，那么对于第 $i$ 个事件的贡献为 $a[i]-(b[i]+c[i])$，根据贪心的策略，我们可以考虑将所有时间按照贡献进行排序，贡献越大则优先选择。

$Y,Z$ 获胜同理。

注意

• $X$ 的获胜条件为 $X>Y+Z$，非 $X\ge Y+Z$。
• 无获胜国时，输出 -1。

AC_Code

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>using namespace std;typedef long long LL;const int N = 1e5 + 10;int n;
int a[N], b[N], c[N], d[N];int get(int a[], int b[], int c[])
{for (int i = 0; i < n; ++ i )d[i] = a[i] - b[i] - c[i];sort(d, d + n, greater<int>());LL res = 0;for (int i = 0; i < n; ++ i ){res += d[i];if (res <= 0)return i;}return n;
}int main()
{cin >> n;for (int i = 0; i < n; ++ i )cin >> a[i];for (int i = 0; i < n; ++ i )cin >> b[i];for (int i = 0; i < n; ++ i )cin >> c[i];int res = max({get(a, b, c), get(b, a, c), get(c, a, b)});if (res == 0)res = -1;cout << res << endl;return 0;
}

---

#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;typedef long long LL;const int N = 1e5 + 10;int n;
struct Node
{int x, y, z;
}q[N];bool cmp_x(Node a, Node b)
{return a.x - a.y - a.z > b.x - b.y - b.z;
}bool cmp_y(Node a, Node b)
{return a.y - a.x - a.z > b.y - b.x - b.z;
}bool cmp_z(Node a, Node b)
{return a.z - a.x - a.y > b.z - b.x - b.y;
}int main()
{cin >> n;for (int i = 0; i < n; ++ i )cin >> q[i].x;for (int i = 0; i < n; ++ i )cin >> q[i].y;for (int i = 0; i < n; ++ i )cin >> q[i].z;LL res = -1, sum = 0, t = 0;sort(q, q + n, cmp_x);for (int i = 0; i < n; ++ i ){sum += q[i].x - q[i].y - q[i].z;if (sum <= 0)break;t ++;}res = max(res, t);sum = 0, t = 0;sort(q, q + n, cmp_y);for (int i = 0; i < n; ++ i ){sum += q[i].y - q[i].x - q[i].z;if (sum <= 0)break;t ++;}res = max(res, t);sum = 0, t = 0;sort(q, q + n, cmp_z);for (int i = 0; i < n; ++ i ){sum += q[i].z - q[i].x - q[i].y;if (sum <= 0)break;t ++;}res = max(res, t);cout << (!res? -1: res) << endl;return 0;
}

【在线测评】