AtCoder Beginner Contest 257

A - A to Z String 2

签到

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

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

int main()
{
    int n , x;
    cin >> n >> x;
    for( char i = 'A' ; i <= 'Z' ; i ++ )
    {
        if( x > n ) x -= n;
        else printf("%c\n" , i ) , exit(0);
    }
}

B - 1D Pawn

判断一下被移动的时候和下一位相邻，是否是边界。然后移动就好了

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


int n , a[205] , k , q;

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

int main()
{
    cin >> n >> k >> q;
    for( int i = 1 ; i <= k ; i ++ )
        cin >> a[i];
    for( int i = 1 , x ; i <= q ; i ++ )
    {
        cin >> x;
        if( a[x] == n ) continue;
        if( a[x] + 1 == a[x+1] ) continue;
        a[x] ++;
    }
    for( int i = 1 ; i <= k ; i ++ )
        cout << a[i] << " ";
    cout << endl;
}

C - Robot Takahashi

这道题的意思是机器人会单纯的按照体重把人分为大人和儿童。现在给出 n 个人的体重和是大人还是小孩。然后问如何界定标准最准确。

首先要知道的是，标准界定在人与人之间还是界定在人是没有影响的，所以我们统计全部的大人数量，然后把所有人排序，并界定为大人，然后依次把每个人变成儿童，在枚举的过程中找到最准确的答案

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

const int N = 2e5+5;
int n , sum , cnt , ans;
pair< int , int > w[N];
string s;


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

int main()
{
    n = read();
    cin >> s;
    for( auto it : s )
        sum += it == '1';
    for( int x , i = 1 ; i <= n ; i ++ )
        w[i].first = read() , w[i].second = s[i-1] == '1';
    sort( w + 1 , w + 1 + n );
    ans = sum;
    for( int i = 1 ; i <= n ; i ++ ){
        if( w[i].second == 0 ) cnt ++;
        else sum --;
        if( w[i].first == w[i+1].first ) continue;
        ans = max( ans , cnt + sum );
    }
    cout << ans << endl;
}

E - Addition and Multiplication 2

这道题其实很简单，先用最便宜的占位，如果有多个最便宜的就用他们之中最大一个，然后从高位开始把每一位尽可能的变大就好了。

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

const int N = 1e6+5;
int n , ans[N] , c[15] , t , p , m ;

int main()
{
    cin >> n;
    for( int i = 1 ; i <= 9 ; i ++ )
        cin >> c[i];
    t = c[9] , p = 9;
    for( int i = 8 ; i >= 1 ; i -- )
        if( c[i] < t ) t = c[i] , p = i;
    m = n / t , n %= t;
    for( int i = 1 ; i <= 9 ; i ++ ) c[i] -= t;
    for( int i = 1 ; i <= m ; i ++ )
    {
        if( n == 0 ) {
            cout << p;
            continue;
        }
        for( int i = 9 ; i >= p ; i -- )
            if( c[i] <= n ) {
                cout << i;
                n -= c[i];
                break;
            }
    }
}

D - Jumping Takahashi 2

这道题我读错了题目，S值在跳的过程中是不会变化的，题目说跳增加S指的是开始时原地跳跃

知道了正确的题意就简单了，我们只要二分答案就好了，判断，我这里采用的是比较笨的方法，就是暴力的跑 dfs，但是也有简单方法就是Floyd 就好

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

const int N = 205 , inf = 5e9;
int n ,sx[N] , sy[N] , sp[N];
vector< int > e[N];
bitset<N> vis;

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 getdist( int u , int v ){
    return abs( sx[u] - sx[v] ) + abs( sy[u] - sy[v] );

}

void dfs( int u ){
    if( vis[u] ) return;
    vis[u] = 1;
    for( auto v : e[u] )
        dfs( v );
    return;
}

bool pending( int v ){
    for( int i = 1 ; i <= n ; i ++ ) e[i].clear();
    vis.reset();
    for( int i = 1 ; i <= n ; i ++ )
        for( int j = 1 ; j <= n ; j ++ )
            if( i == j ) continue;
            else {
                if( getdist( i , j ) <= sp[i] * v ) e[i].push_back(j);
            }
    for( int i = 1 ; i <= n ; i ++ ){
        vis.reset();
        dfs( i );
        bool flag = 1;
        for( int j = 1 ; j <= n ; j ++ )
            if( !vis[j] ) flag = 0;
        if( flag ) return 1;
    }
    return 0;
}

int32_t main() {
    n = read();
    for( int i = 1 ; i <= n ; i ++ )
        sx[i] = read() , sy[i] = read() , sp[i] = read();

    int l = 1 , r = inf , mid , ans ;

    while( l <= r ){
        mid = ( l + r ) >> 1;
        if( pending( mid ) ) ans = mid , r = mid - 1 ;
        else l = mid + 1;
    }
    cout << ans << endl;
}