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;
}