西南民族大学 春季 2023 训练赛 6
题目详情 - L1-1 今天我要赢 (pintia.cn)
void solve() { cout << "I'm gonna win! Today!" << endl; cout << "2022-04-23" << endl; }
题目详情 - L1-2 种钻石 (pintia.cn)
void solve() { int v; cin >> n >> v; cout << n / v << endl; }
题目详情 - L1-3 谁能进图书馆 (pintia.cn)
int n,m,t,sa,pa,q1,q2; void solve() { cin >> sa >> pa >> q1 >> q2 ; if(q1 >= sa && q2 >= sa) { cout << q1 << "-Y " << q2 << "-Y" << endl; cout << "huan ying ru guan" << endl; } else if(q1 >= pa) { cout << q1 << "-Y " << q2 << "-Y" << endl; cout << "qing 1 zhao gu hao 2" << endl; } else if(q2 >= pa) { cout << q1 << "-Y " << q2 << "-Y" << endl; cout << "qing 2 zhao gu hao 1" << endl; } else if(q1 >= sa) { cout << q1 << "-Y " << q2 << "-N" << endl; cout << "1: huan ying ru guan" << endl; } else if(q2 >= sa) { cout << q1 << "-N " << q2 << "-Y" << endl; cout << "2: huan ying ru guan" << endl; } else { cout << q1 << "-N " << q2 << "-N" << endl; cout << "zhang da zai lai ba" << endl; } }
题目详情 - L1-4 拯救外星人 (pintia.cn)
int n,m,t,sa,pa,q1,q2; void solve() { cin >> n >> m; int ans = 1; for(int i = 1; i <= n +m; i++) ans *= i; cout << ans << endl; }
题目详情 - L1-5 试试手气 (pintia.cn)
int n,m,t,sa,pa,q1,q2; int a[6]; void solve() { for(int i = 0; i <6 ;i ++) cin >> a[i]; cin >> n; for(int j = 0; j < 6;j ++) { int d = n; for(int i = 6; i > 0; i--) { if(i == a[j]) continue; d--; if(d == 0) { cout << i << (j == 5 ? "\n" : " "); break; } } } }
题目详情 - L1-6 斯德哥尔摩火车上的题 (pintia.cn)
void solve() { string s1,s2,ans1 = "", ans2 = ""; cin >> s1 >> s2; for (int i = 1; i < s1.size(); i++) { if (s1[i] % 2 == s1[i-1] % 2) { ans1 += max(s1[i], s1[i-1]); } } for (int i = 1; i < s2.size(); i++) { if (s2[i] % 2 == s2[i-1] % 2) { ans2 += max(s2[i], s2[i-1]); } } if(ans1 == ans2) cout << ans1 << endl; else { cout << ans1 << endl << ans2 << endl; } }
题目详情 - L1-7 机工士姆斯塔迪奥 (pintia.cn)
总格数就相当于把BOSS攻击的行和列放在最边上,最后计算被截下的长方形格数即可,BOSS会攻击已经攻击过得行和列,所以要判定该行/列是否被攻击过
int hang[N], lie[N]; void solve() { cin >> n >> m >> q; int ans = n * m; int d = q; while(q--) { int c; cin >> t >> c; if(t) { if(!lie[c]) m--; lie[c] = 1; } else { if(!hang[c]) n--; hang[c] = 1; } } cout << n * m << endl; }
题目详情 - L1-8 静静的推荐 (pintia.cn)
#include <iostream> #include <string.h> #include <stdio.h> #include <vector> #include <map> #include <queue> #include <algorithm> #include <math.h> #include <cstdio> #include <set> #include <utility> #define inf 0x3f3f3f3f #define endl '\n' #define int long long #define f first #define s second using namespace std; const int N = 1e5+10, M = 1e6 + 10; //typedef long long ll; typedef pair<int,int> PII; //queue<PII> q1; //priority_queue <int,vector<int>,greater<int> > q2; int n,m,t,sa,pa,q,k,sorce,pat[N]; /* */ struct Sorce{ int ss,pat; bool operator < (const Sorce &s) const{ if(s.ss != ss) return pat < s.pat; return ss < s.ss; } }; vector<Sorce> sr(N); bool vis[N]; int enough[N]; void solve() { cin >> n >> k >> sorce; int ans = 0; for(int i = 0; i < n; i++) { int x,y; cin >> x >> y; if(x >= 175 && y >= sorce) { ans ++; continue; } sr.push_back({x, y}); //cin >> sorce[i] >> pat[i]; } sort(sr.begin(), sr.end()); t = 0; for(int i = 0 ; i < sr.size(); i++) { if(sr[i].ss < 175) continue; t = max(t, sr[i].ss); enough[sr[i].ss - 175] ++; } t -= 175; for(int i = 0; i <= t; i++) { //cout << enough[i] << ' '; if(enough[i] >= k) ans += k; else ans += enough[i]; } cout << ans << endl; } signed main() { ios::sync_with_stdio(false); cin.tie(0);cout.tie(0); int Ke_scholar = 1; while(Ke_scholar--) { solve(); } return 0; }
题目详情 - L2-1 插松枝 (pintia.cn)
题目详情 - L2-2 老板的作息表 (pintia.cn)
将输入的时间换算成秒进行排序,如果这段区间的起始时间和上一段的结束时间相等就跳过,否则就输出
#include <iostream> #include <string.h> #include <stdio.h> #include <vector> #include <map> #include <queue> #include <algorithm> #include <math.h> #include <cstdio> #include <set> #include <utility> #define inf 0x3f3f3f3f #define endl '\n' #define int long long #define f first #define s second using namespace std; const int N = 1e6+10, M = 1e6 + 10; //typedef long long ll; typedef pair<int,int> PII; //queue<PII> q1; //priority_queue <int,vector<int>,greater<int> > q2; int n,m,t,sa,pa,q,k,sorce,pat[N]; /* */ bool vis[N],st[N]; vector<PII> a; void solve() { cin >> n; a.push_back({0,0}); a.push_back({24 * 60 * 60 - 1, 24 * 60 * 60 - 1}); while(n--) { char c1, c2, c3, c4, c5; int h1, h2, m1, m2, s1, s2; cin >> h1 >> c1 >> m1 >> c2 >> s1 >> c3 >> h2 >> c4 >> m2 >> c5 >> s2; a.push_back({(h1 * 60 + m1) * 60 + s1, (h2 * 60 + m2) * 60 + s2}); } sort(a.begin(), a.end()); for(int i = 1;i < a.size(); i ++) { if(a[i].f == a[i - 1].s) continue; int h = a[i].f / (60 * 60), m = a[i].f % 3600 / 60, s = a[i].f % 60; int h0 = a[i - 1].s / (60 * 60), m0 = a[i - 1].s % 3600 / 60, s0 = a[i - 1].s % 60; printf("%02d:%02d:%02d - ",h0,m0,s0); printf("%02d:%02d:%02d\n",h,m,s); } } signed main() { ios::sync_with_stdio(false); cin.tie(0);cout.tie(0); int Ke_scholar = 1; while(Ke_scholar--) { solve(); } return 0; }
