题目链接
A
4个相等的数异或为0，直接输出0
B
设正数数量为 a，负数数量为 b。当 a = b，res = a + b
当 a != b，那么相当于将多的放在最前面和最后面，res = 2 * min(a, b) + 1;
C
求同行和同列的最大极差
#include <iostream>
#include <vector>
using namespace std;
const int N = 1e5 + 10;
vector<int> row[N], col[N];
int n, m;int main()
{cin >> n >> m; while (m--){int x, y; cin >> x >> y;row[x].push_back(y), col[y].push_back(x);}int res = 0;for (int i = 1; i <= n; i++){if (row[i].size() < 2) continue;int mx = -1, mi = 1e6;for (auto& e : row[i])mx = max(mx, e), mi = min(mi, e);res = max(res, mx - mi);}for (int i = 1; i <= n; i++){if (col[i].size() < 2) continue;int mx = -1, mi = 1e6;for (auto& e : col[i])mx = max(mx, e), mi = min(mi, e);res = max(res, mx - mi);}cout << res << endl;
}
D
维护一个集合（sum），表示当前可以组成的数的上界。先将数组排序，比较当前数 a[i] 和上界的大小：
- a[i] > sum + 1，表示 sum + 1 无法被构造出来
- a[i] <= sum + 1，当前可以构成的上界为：sum + a[i]
using LL = long long;
const int N = 1e5 + 10;
int a[N], n;void slove()
{cin >> n;for (int i = 1; i <= n; i++) cin >> a[i];sort(a + 1, a + n + 1);LL sum = 0;for (int i = 1; i <= n; i++) {if (a[i] <= sum + 1) {sum += a[i];}}if (sum >= n) cout << "Cool!" << endl;else cout << sum + 1 << endl;
}
E
一个组合数，中间长度为 n - 2，能推 m - 1 次，问有多少种推法？
C(n - 2, m - 1)
#include<iostream>
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
const int Mod = 1e9 + 7;
int fact[N], infact[N];
int n, m;
int qmi(int a, int k, int p)
{int res = 1;while(k){if(k & 1) res = (ll) res * a % p;a = (ll) a * a % p;k >>= 1;}return res;
}void init()
{infact[0] = fact[0] = 1;for(int i = 1; i < N; i++)fact[i] = (ll) fact[i - 1] * i % Mod;for(int i = 1; i < N; i++)infact[i] = (ll) infact[i - 1] * qmi(i, Mod - 2, Mod) % Mod;
}int main()
{int t; cin >> t;init();while(t--){cin >> n >> m;int a = n - 2, b = m - 1;cout << (ll) fact[a] * infact[b] % Mod * infact[a - b] % Mod << endl;}return 0;
}