蓝桥杯选做

杂项

错误票据【第四届】【省赛】

#include <bits/stdc++.h>
using namespace std;

int n;
vector<int> a;


int main()
{
    int n , m ;
    string s;
    cin >> n;
    getchar();
    for( int i = 1 ; i <= n  ; i ++ )
    {
        getline( cin , s ) , m = 0;
        for( int t = 0; t <= s.size() - 1 ; t ++  )
        {
            if( s[t] == ' ' )
            {
                a.push_back( m );
                m = 0;
            }
            else m = m * 10 + s[t] - '0';
        }
        a.push_back(m);
    }
    sort(  a.begin() , a.end() );

    for( int i = 0 ; i < a.size() - 1 ; i ++ )
    {

        if( a[i] + 1 != a[ i + 1 ] && a[i] != a[ i + 1 ] ) n = a[ i ] + 1;
        if( a[i] == a[ i + 1 ] ) m = a[i];
    }
    cout << n << " " << m << endl;
    return 0;
}

翻硬币【第四届】【省赛】

#include <bits/stdc++.h>
using namespace std;

int cnt;
string a , b;

int main()
{
    cin >> a >> b;
    for( int i = 0 ; i < a.size() ; i ++ )
    {
        if( a[i] == b[i] ) continue;
        cnt ++;
        a[i] = ( a[i] == 'o' ? '*' : 'o' );
        a[ i + 1 ] = ( a[ i + 1 ] == 'o' ? '*' : 'o' );
    }
    cout << cnt << endl;
    return 0;
}	

2020年C++省赛 B 组

门牌制作

624

即约分数

2481215

蛇形填数

打表代码

#include <bits/stdc++.h>
using namespace std;

const int N = 50;
int a[N][N];

int main(){
    int k = 1;

    for( int i = 1 ; i <= 50 ; i ++ )
    {
        if( i & 1)
            for( int j = 1 ; j <= i ; j ++ )
                a[i][j] = k , k ++;
        else
            for( int j = i ; j >= 1 ; j -- )
                a[i][j] = k , k ++;
    }
    for( int i = 1 ; i <= 5 ; i ++ ) {
        for (int j = 1; j <= 5 ; j++)
            printf("%4d" , a[i+j-1][j] );
        cout << endl;
    }
    cout << a[39][20] << endl;
}

答案

761

七段码

打表代码

就是用 dfs 枚举出所有的情况,再用并查集判断是否在同一个集合中

#include <bits/stdc++.h>
using namespace std;

const int N = 10;
int res, fa[N];
bool e[N][N] , use[N];

int getfa( int x )
{
    if( fa[x] == x ) return x;
    return fa[x] = getfa( fa[x] );
}

void merge( int x , int y )
{
    x = getfa(x) , y = getfa(y);
    fa[x] = y;
    return ;
}

void dfs( int d )
{
    if( d > 7 ){
        for( int i = 1 ; i <= 7 ; i ++ )
            fa[i] = i;
        for( int i = 1 ; i <= 7 ; i ++ )
            for( int j = 1 ; j <= 7 ; j ++ )
                if( e[i][j] && use[i] && use[j] )
                    merge( i , j );
        int k = 0;
        for( int i = 1 ; i <= 7 ; i ++ )
            if( use[i] && fa[i] == i ) k ++;
        if( k == 1 ) res ++;
        return;
    }
    dfs(d+1);
    use[d] = 1;
    dfs(d+1);
    use[d] = 0;
    return ;
}

int main()
{
    e[1][2] = e[1][6] = 1;
    e[2][1] = e[2][3] = e[2][7] = 1;
    e[3][7] = e[3][4] = e[3][2] = 1;
    e[4][3] = e[4][5] = 1;
    e[5][4] = e[5][6] = e[5][7] = 1;
    e[6][1] = e[6][7] = e[6][1] = 1;
    dfs(1);
    cout << res << endl;
}

跑步锻炼

8879

打表代码

year = 2000
month = 1
date = 1
cnt = 0
week = 6

while 1 :
    cnt += 1
    if( date == 1 or week == 1 ):
        cnt += 1

    if year == 2020 and month == 10 and date == 1:
        break

    week += 1
    if week == 8 :
        week = 1

    date += 1
    if ( month == 1 or month == 3 or month == 5 or month == 7 or month == 8 or month == 10 or month == 12 ) and date == 32:
        date = 1
        month += 1
    elif ( year == 2000 or year == 2004 or year == 2008 or year == 2012 or year == 2016 or year == 2020 ) and month == 2 and date == 30 :
        month += 1
        date = 1
    elif not ( year == 2000 or year == 2004 or year == 2008 or year == 2012 or year == 2016 or year == 2020 ) and month == 2 and date == 29 :
        date = 1
        month += 1
    elif not( month == 1 or month == 3 or month == 5 or month == 7 or month == 8 or month == 10 or month == 12 or month == 2 ) and date == 31:
        date = 1
        month += 1

    if month == 13 :
        month = 1
        year += 1

print(cnt)

成绩统计

签到题

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int n , a , b;

int main()
{
    cin >> n;
    for( int i = 1 , x ; i <= n ; i ++ )
    {
        cin >> x ;
        if( x >= 60 ) a ++;
        if( x >= 85 ) b ++;
    }
    printf("%.0lf%%\n%.0lf%%\n" , 100.0*a/n , 100.0*b/n );

}

子串分值和

60 分

就是枚举起点和终点,然后统计一下就好

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5+5 , M = 30;
int n , res , cnt ;
string s;
bool vis[M];

int main()
{
    cin >>s;
    n = s.size();
    for( int i = 0 ; i < n ; i ++ )
    {
        memset( vis , 0 , sizeof(vis) );
        cnt = 0;
        for( int j = i , t ; j < n ; j ++ )
        {
            t = s[j] - 'a';
            if( !vis[t] )
                cnt ++ , vis[t] = 1;
            res += cnt;
        }
    }
    cout << res << endl;
}

100分

在上面的基础上,如果[i,j]有 26 个字母那么[i,j+1]也是 26 个后面的每一个都是 26 个没必要继续循环下去

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int N = 1e5+5 , M = 30;
int n ;
ll res , cnt ;
string s;
bool vis[M];

int main()
{
    cin >>s;
    n = s.size();
    for( int i = 0 ; i < n ; i ++ )
    {
        memset( vis , 0 , sizeof(vis) );
        cnt = 0;
        for( int j = i , t ; j < n ; j ++ )
        {
            t = s[j] - 'a';
            if( !vis[t] )
                cnt ++ , vis[t] = 1;
            if( cnt == 26 )
            {
                res += ( n - j ) * cnt;
                break;
            }
            res += cnt;
        }
    }
    cout << res << endl;
}

但是这个做法没有通过 acw 的加强数据

正解

就是 dp , f[i]表示右端点为i的所有区间的和,如果s[i+1]在1到i中最后一次出现的位置是p那么f[i+1]=f[i]+i-p,因为 p之后的每一个区间都要加一

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int N = 1e5+5 , M = 30;
int n , pos[N];
ll res , f[N];
char s[N];

int main()
{
    scanf("%s" , s+1 );
    n = strlen(s+1);
    for( int i = 1 , t; i <= n ; i ++ )
        t = s[i] - 'a' , f[i] = f[i-1] + i - pos[t] , pos[t] = i , res += f[i];
    cout << res << endl;
}

回文日期

这题就是首先向枚举日期,再进行判断

这里有一个问题是,b 一定是 a,a 不一定是 b 且,b不等于 a,所以可以推断 b>a,证明可以用反证法

#include <bits/stdc++.h>
#define ll long long
using namespace std;

int n;
int months[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

bool check_date( int x )
{
    int year = x/10000 , month = x%10000/100 , day = x%100;
    if(!day || month < 1 || month > 12 ) return 0;
    if( month != 2 && day > months[month] ) return 0;
    if( month == 2 ){
        if( year%4==0 && year%100!=0 || ( year%400==0) ) {
            if(day>29) return 0;
        }
        else{
            if(day>28) return 0;
        }
    }
    return 1;
}
bool check1(string s){
    int len = s.size();
    for( int i = 0 , j = len - 1 ; i < j ; i ++ , j -- )
        if( s[i] != s[j] ) return 0;
    return 1;
}

bool check2(string s){
    if(!check1(s)) return 0;
    if( s[0]!= s[2] || s[1]!=s[3] || s[0] == s[1] ) return 0;
    return 1;
}

void slove()
{
    cin >> n;
    for( int i = n+1 , flag = 0 ; ; i ++ )
    {
        if( !check_date(i) ) continue;
        string s = to_string(i);
        if( check1(s) && !flag )
        {
            cout << i << endl;
            flag = 1;
        }
        if(check2(s))
        {
            cout << i << endl;
            return ;
        }

    }
}
int main()
{
    int t;
    cin >> t;
    while( t-- )
        slove();
    return 0;
}

平面切分

每增加一条线,区域数就会加一,当前线与之前每多一个交点区域数也会加一

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int N = 1005;
int n , m , cnt = 1;
pair< int , int > e[N];
set< pair<double,double> > s;

int read(){
    int x = 0 , f = 1 , ch = getchar();
    while( (ch<'0'||ch>'9') && ch != '-' ) ch = getchar();
    if( ch == '-' ) f = -1 , ch = getchar();
    while( ch >= '0' && ch <= '9' ) x = (x<<3)+(x<<1) + ch-'0' , ch = getchar();
    return x*f;
}

int main()
{
    m = n = read();
    for( int i = 1 ; i <= n ; i ++ )
        e[i].first = read() , e[i].second = read();
    sort( e+1 , e+1+n );
    n = unique( e+1 , e+1+n ) - e - 1;
    double nx , ny ;
    for( int i = 1 ; i <= n ; i ++ ){
        s.clear() , cnt ++;
        for( int j = 1 ; j < i ; j ++ )
            if( e[i].first != e[j].first )
                nx = 1.0*(e[i].second-e[j].second)/(e[j].first-e[i].first) , ny = e[i].first*nx+e[i].second , s.insert({nx,ny});
        cnt += s.size();
    }
    cout << cnt << endl;
}