防线
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int N = 200010; 4 typedef long long ll; 5 struct seq { 6 int s, e, d; 7 } seqs[N]; 8 int n; 9 ll get_sum(int x) { //get_sum返回小于等于x的数的个数 10 ll res = 0; 11 for (int i = 0; i < n; i++) { //遍历所有序列 12 if (seqs[i].s <= x) { 13 res += (min(seqs[i].e, x) - seqs[i].s) / seqs[i].d + 1; 14 } 15 } 16 return res; 17 } 18 int main() { 19 ios::sync_with_stdio(false); 20 cin.tie(0); 21 cout.tie(0); 22 int t; 23 cin >> t; 24 while (t--) { 25 int l = 0, r = 0; 26 cin >> n; 27 for (int i = 0; i < n; i++) { 28 int s, e, d; 29 cin >> s >> e >> d; 30 seqs[i] = {s, e, d}; 31 r = max(r, e); 32 } 33 while (l < r) { 34 int mid = (l + r) / 2; 35 if (get_sum(mid) & 1) { //如果是奇数的话 36 r = mid; 37 } else { 38 l = mid + 1; 39 } 40 } 41 //看一下二分出来的这个点上的数的个数是多少 42 //小于等于l的数的个数减去小于等于l - 1的数的个数就是等于l的数的个数 43 ll sum = get_sum(l) - get_sum(l - 1); 44 if (sum & 1) { //如果是奇数，就是答案 45 cout << l << " " << sum << endl; 46 } else { //否则无解 47 cout << "There's no weakness." << endl; 48 } 49 } 50 return 0; 51 }