PTA天梯赛练习集选做

002

模拟+找规律

规律就是上面正三角第i行是i*2-1个符号

正三角的符号数为t，整个沙漏的符号数为t*2-1

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


int n , m , t = 0;
char ch;


int main()
{
    cin >> n >> ch;

    for( register int i = 1 ; ; i ++ )
    {
        m = i * 2 - 1;
        t += m;
        if( ( t * 2 - 1) <= n ) continue;
        t -= m;
        m = i - 1;
        break;
    }
    for( register int i = m ; i >= 1 ; i -- )
    {
        for( register int j = 1 ; j <= m - i ; j ++ ) putchar(' ');
        for( register int j = 1 ; j <= i * 2 - 1 ; j ++ ) putchar(ch);
        puts("");
    }
    for( register int i = 2 ; i <= m ; i ++ )
    {
        for( register int j = 1 ; j <= m - i ; j ++ ) putchar(' ');
        for( register int j = 1 ; j <= i * 2 - 1 ; j ++ ) putchar(ch);
        puts("");
    }
    cout << n - ( t * 2 - 1) << endl;
    return 0;
}

003

很简单，数字太长用sting存，用auto便历

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

int t[15];

int main()
{
    string s;
    cin >> s;
    for( auto it : s )
    {
        t[ it - '0' ] ++;
    }
    for( int i = 0 ; i <= 9 ; i ++ )
    {
        if( t[i] ) printf( "%d:%d\n" , i , t[i] );
    }
    return 0;
}

004

printf("Celsius = %d\n" ,  5 * ( a -32 ) / 9 );

005

这道题告诉我们要活用STL，map套pair可以很简单

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


map < int , pair< string , int > > t;
int  n , m;
string s;

int main()
{
    cin >> n;
    for( int i = 1 , a , b; i <= n ; i ++ )
    {
        cin >> s >> a >> b ;
        t[a] = make_pair( s , b );
    }
    cin >> m;
    for( int i = 1 , a ; i <= m ; i ++ )
    {
        cin >> a;
        cout << t[a].first << ' ' << t[a].second << endl;
    }
    return 0;
}

006

一道挺有意思的枚举题，分别枚举左右短点就行

在配合一点优化就行

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

long long n , t , maxlong = 0 , l, r , mul = 1 , temp;

int main()
{
    cin >> n;
    t = sqrt( n ) + 1, l = n , r = n;
    for( register int i = 2 ;i <= t ; i ++ )//枚举左端点
    {
        if( n % i ) continue;
        temp = 1;
        for( register int j = i ; j < i + maxlong ; j ++ ) temp *= j;
        for( register int j = i + maxlong ; j <= t ; j ++ )//枚举右端点
        {
            temp *= j;
            if( temp > n ) break;
            if( n % temp ) break;
            l = i , r = j , maxlong = j - i + 1;
        }
    }
    if( maxlong == 0 ) maxlong = 1;
    cout << maxlong << endl;
    for( register int i = l ; i < r ; i ++ ) cout << i << '*';
    cout << r << endl;
    return 0;
}