菜狗现在才开始备战蓝桥杯QAQ

文章目录

• 【蓝桥杯专题】 （C++ | 洛谷 | acwing | 蓝桥）
• 回文日期
• 纸张尺寸 蓝桥杯真题
• 错误票据
• AcWing 788. 逆序对的数量
• 航班时间
• 移动距离
• 连号区间
• 1236. 递增三元组
• P
• P
• P
• P
• P

【蓝桥杯专题】 （C++ | 洛谷 | acwing | 蓝桥）

文章目录

• 【蓝桥杯专题】 （C++ | 洛谷 | acwing | 蓝桥）
• 回文日期
• 纸张尺寸 蓝桥杯真题
• 错误票据
• AcWing 788. 逆序对的数量
• 航班时间
• 移动距离
• 连号区间
• 1236. 递增三元组
• P
• P
• P
• P
• P

回文日期

链接 链接

#include 
// #include 
using namespace std;
typedef long long ll;
typedef double db;
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define per(i, a, n) for(int i = n; i <= a; i --)
#define pb push_back;
#define fs first;
#define sz second;
#include  // atoi
#define debug cout<<"debug"<<"\n"
#define endl "\n";
const int INF = 0x3f3f3f3f;
const int mod=1e9+7;
const int N = 1e5 + 10;int months[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};// int months[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool check (int date) {int year = date / 10000;int month = date % 10000 / 100;int day = date % 100;//判断 离谱日期if(month > 13 || !day || !month) return false;// 判断月份对应日期if(month != 2 && day > months[month]) return false;//判断润年if(month == 2) {//如果能被 400 整除  并且 能被四整除 或不能被100整除bool leap = year % 400 == 0 &&  year % 4 == 0 || year % 100;if(day > leap + 28) return false;}return true;
}void solve () {int s1, s2;cin >> s1 >> s2;int res = 0;    for(int i = 1; i < 10000; i ++) {int x = i, r = i;for(int j = 0; j < 4; j ++) {r = r * 10 + x % 10, x /= 10;}if(r >= s1 &&  r <= s2 && check(r)) res ++;}cout << res << endl;
}int main(void){
// 	freopen("in.txt","r",stdin);ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;// cin >> T;while(T --) solve();return 0;
}

纸张尺寸 蓝桥杯真题

• 模拟题
  scanf 新操作
  

错误票据

链接 链接

• 读入 可以学一下

#include 
// #include 
using namespace std;
typedef long long ll;
typedef double db;
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define per(i, a, n) for(int i = n; i <= a; i --)
#define pb push_back;
#define fs first;
#define sz second;
#include  // atoi
#define debug cout<<"debug"<<"\n"
#define endl "\n";
const int INF = 0x3f3f3f3f;
const int mod=1e9+7;
const int N = 1e5 + 10;int a[N];
void solve () {string line;int n = 0;int cnt;cin >> cnt;getline(cin,line);while (cnt --) { // 空间多开点  因为很大 80 一行可能10个 就是 800 甚至更多getline(cin,line);stringstream ssin(line);while(ssin >> a[n]) n ++;}sort(a,a + n + 1);int res1, res2;for(int i = 0; i <= n; i ++) {if(a[i] == a[i - 1] + 2)res1 = a[i - 1] + 1;if(a[i] == a[i - 1]) res2 = a[i];}cout << res1 << ' ' << res2 << endl;
}int main(void){freopen("in.txt","r",stdin);     ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;// cin >> T;while(T --) solve();return 0;
}

AcWing 788. 逆序对的数量

链接 链接

//归并排序#include 
// #include 
using namespace std;
typedef long long ll;
typedef double db;
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define per(i, a, n) for(int i = n; i <= a; i --)
#define pb push_back;
#define fs first;
#define sz second;
#include  // atoi
#define debug cout<<"debug"<<"\n"
#define endl "\n";
const int INF = 0x3f3f3f3f;
const int mod=1e9+7;
const int N = 1e5 + 10;ll res = 0;
int q[N], w[N];void merge_sort(int q[], int l, int r) {    if(l >= r) return;int mid = l + r >> 1;merge_sort(q, l, mid) ,merge_sort(q, mid + 1, r);int i = l, j = mid + 1, k = 0;while(i <= mid && j <= r) {if(q[i] <= q[j]) w[k ++] = q[i ++];else {w[k ++] = q[j ++];res += mid - i + 1; // 与递归的区别 ， 把不同元素区间}} while(i <= mid ) w[k ++] = q[i ++];while(j <= r) w[k ++] = q[j ++];for(i = l, j = 0; i <= r; i ++, j ++) q[i] = w[j];}void solve () {int n;cin >> n;rep(i, 0, n - 1) {cin >> q[i];}merge_sort(q, 0, n - 1);cout << res <
// 	freopen("in.txt","r",stdin);ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;// cin >> T;while(T --) solve();return 0;
}

航班时间

 scanf("%d(+%d)%d", &a, &b, &c);   如果不匹配 直接忽略 b的读

• 图
  

链接 链接

#include
#include
#include
#include 
#include
#include 
#include 
using namespace std;
typedef long long ll;
typedef double db;
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define per(i, a, n) for(int i = n; i <= a; i --)
#define pb push_back;
#define fs first;
#define sz second;
#include  // atoi
#define debug cout<<"debug"<<"\n"
#define endl "\n";
const int INF = 0x3f3f3f3f;
const int mod=1e9+7;
const int N = 1e5 + 10;int getTime(void)
{int h1,m1,s1,h2,m2,s2,d=0;scanf("%d:%d:%d %d:%d:%d (+%d)",&h1,&m1,&s1,&h2,&m2,&s2,&d);int time=d*24*3600+h2*3600+m2*60+s2-(h1*3600+m1*60+s1);return time;
}void solve () {int t;scanf("%d",&t);for(int i = 0; i < t; i++){int time1=getTime();int time2=getTime();int t=(time1+time2)/2;printf("%02d:%02d:%02d\n", t/3600, t/60%60, t%60);}
}int main(void){freopen("in.txt","r",stdin);ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;cin >> T;while(T --) solve();return 0;
}

移动距离

链接 链接

• 简单数学模拟 + 类似蛇形填数

#include
#include
#include
#include 
#include
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
typedef double db;
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define per(i, a, n) for(int i = n; i <= a; i --)
#define pb push_back;
#define fs first;
#define sz second;
#include  // atoi
#define debug cout<<"debug"<<"\n"
#define endl "\n";
const int INF = 0x3f3f3f3f;
const int mod=1e9+7;
const int N = 1e5 + 10;void solve () {int w, n, m;cin >> w >> n >> m;m -- , n --;int x1 = m % w;int y1 = m / w;int x2 = n % w;int y2 = n / w;if(y1 % 2) x1 = w - x1 - 1;if(y2 % 2) x2 = w - x2 - 1;int ans = abs(x2 - x1) + abs(y2 - y1);cout << ans << endl;
}int main(void){freopen("in.txt","r",stdin);ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;// cin >> T;while(T --) solve();return 0;
}

链接 链接

连号区间

• 思路 ： 区间连续 则 边界差 == max - min
  

链接 链接

#include 
#include 
#include using namespace std;int a[10010];int main () {int n;cin >> n;for(int i = 0; i < n; i ++) {cin >> a[i];}long long res = 0;//因为是连续的所以在所取的[l，r]范围中寻找最大值，// 最小值。然后相减，最后和r-l（区间长度）作比较即可。 // 范围 i ~ jfor(int i = 0; i < n; i ++) {int minn = 10000;int maxx = -10000;for(int j = i; j < n; j ++ ) {maxx = max(maxx, a[j]); // 更新区间值minn = min(minn, a[j]);if((j - i)  == (maxx - minn)) res ++;}}cout <

#include 
#include 
#include using namespace std;const int N = 100010;
int a[N], b[N], c[N];int main () {int n;long long res = 0;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];sort(a, a + n);sort(b, b + n);sort(c, c + n);for(int i = 0; i < n; i ++) {int x = b[i];int l = 0, r = n - 1;int aa = 0, bb = 0;while(l < r) {int mid = l + r + 1 >> 1;if(a[mid] < x) l = mid;else r = mid - 1;}//二分结果未必找得到  所以需要判断if(a[l] >= b[i]) {r = -1;} aa = r;l = 0, r = n - 1;while(l < r) {int mid = (l + r) / 2;if(c[mid] > x) r = mid;else l = mid + 1;}//二分结果未必找得到  所以需要判断if(b[i] >= c[l]){r = n;} bb = r;res += (long long )(aa + 1) * (n - bb);}cout << res <