Codeforces Round#490 Div3

62 阅读 0 评论 41 点赞

我是靠谱客的博主文艺芹菜，最近开发中收集的这篇文章主要介绍Codeforces Round#490 Div3，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

这次的题解，我想划划水，因为前三题很快就写出来了，而且对了，第四题写了一个不过一直超时，赛后改了一下，也过了。

A题

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
int n,k;
int a[104];
int main()
{
    scanf("%d %d",&n,&k);
    for( int i = 1 ; i <= n ; i++ )
        scanf("%d",&a[i]);
    int l = 1;
    while( a[l] <= k && l <= n )
        l++;
    int r = n;
    while( a[r] <= k && r >= 1 )
        r--;
    int ans = l-1+n-r;
    if( ans > n )
        ans = n;
    cout<<ans<<endl;
    return 0;
}

B题

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
int n,k;
char s[105];
vector<int> v;

void change( int r )
{
    char temp[104];
    int cnt = 1;
    for( int i = r ; i >= 1 ; i-- )
        temp[cnt++] = s[i];
    for( int i = 1 ; i <= r ; i++ )
        s[i] = temp[i];
}
int main()
{
    scanf("%d",&n);
    scanf("%s",s+1);
    for( int i = 2 ; i <= n ; i++ )
        if( n%i == 0 )
            v.push_back(i);
    for( int i = 0 ; i < v.size() ; i++ )
    {
        change(v[i]);
    }
    printf("%sn",s+1);
    return 0;
}

C题

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 5e5+5;
char s[maxn];
int n,k;
int num[30];
int judge[30];
int main()
{
    memset(judge,0,sizeof judge);
    memset(num,0,sizeof num);
    scanf("%d %d",&n,&k);
    scanf("%s",s);
    for( int i = 0 ; i < n ; i++ )
    {
        int t = s[i]-'a';
        num[t]++;
    }
    int rem = k;
    for( int i = 0 ; i < 26 ; i++ )
    {
        if( rem <= num[i] )
        {
            judge[i] = rem;
            break;
        }
        else
        {
            judge[i] = -1;
            rem -= num[i];
        }
    }
    for( int i = 0 ; i < n; i++ )
    {
        int t = s[i]-'a';
        if( judge[t] == 0 )
            printf("%c",s[i]);
        else if( judge[t] > 0 )
            judge[t]--;
    }
    printf("n");
    return 0;
}

D题

You are given an array consisting of $n$ integers $a_{1}, a_{2}, \dots, a_{n}$ , and a positive integer $m$ . It is guaranteed that $m$ is a divisor of $n$ .

In a single move, you can choose any position $i$ between $1$ and $n$ and increase $a_{i}$ by $1$ .

Let's calculate $c_{r}$ ( $0 \leq r \leq m - 1)$ — the number of elements having remainder $r$ when divided by $m$ . In other words, for each remainder, let's find the number of corresponding elements in $a$ with that remainder.

Your task is to change the array in such a way that $c_{0} = c_{1} = \dots = c_{m - 1} = \frac{n}{m}$ .

Find the minimum number of moves to satisfy the above requirement.

Input

The first line of input contains two integers $n$ and $m$ ( $1 \leq n \leq 2 \cdot 10^{5}, 1 \leq m \leq n$ ). It is guaranteed that $m$ is a divisor of $n$ .

The second line of input contains $n$ integers $a_{1}, a_{2}, \dots, a_{n}$ ( $0 \leq a_{i} \leq 10^{9}$ ), the elements of the array.

Output

In the first line, print a single integer — the minimum number of moves required to satisfy the following condition: for each remainder from $0$ to $m - 1$ , the number of elements of the array having this remainder equals $\frac{n}{m}$ .

In the second line, print any array satisfying the condition and can be obtained from the given array with the minimum number of moves. The values of the elements of the resulting array must not exceed $10^{18}$ .

   Examples 
 
     input 
    
      Copy 
    
6 3
3 2 0 6 10 12

     output 
    
      Copy 
    
3
3 2 0 7 10 14 

     input 
    
      Copy 
    
4 2
0 1 2 3

     output 
    
      Copy 
    
0
0 1 2 3

有n个数，和一个除数。

对于这n个数，每个数去除以m，得到的余数进行统计，使0---- m-1 每个都是n/m。

比如 8 4

1 1 2 2 3 3 4 4

余数为0 的要有8/4 = 2个，余数为1的要有2个，余数为2和3的也是。

所以，很显然，我们需要把两个4加1

即，通过增加某个数的值，使得满足条件

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 2e5+5;
int n,m;
ll a[maxn];
int b[maxn];
int num[maxn];
struct spe
{
    int time;
    int add;
};
struct sm
{
    int index,time;
};
vector<spe> ans[maxn];
vector<sm> small;
vector<sm> solve;
int main()
{
    memset(num,0,sizeof num);
    scanf("%d %d",&n,&m);
    for( int i =  1; i <= n ; i++ )
     {
        scanf("%I64d",&a[i]);
        int t = a[i]%m;
        b[i] = t;
        num[t]++;
     }
    int lev = n/m;
    sm temp1;
    spe temp2;
    for( int i = m-1 ; i >= 0 ; i-- )
    {
        if( num[i] < lev )
        {
            temp1.index = i;
            temp1.time = lev-num[i];
            small.push_back(temp1);
        }
        else if( num[i] > lev )
        {
            int t = num[i]-lev;
            while( !small.empty() && t > 0 )
            {
                if( t >= small.back().time )
                {
                    temp2.time = small.back().time;
                    temp2.add = small.back().index-i;
                    ans[i].push_back(temp2);
                    t -= small.back().time;
                    small.pop_back();
                }
                else if( t < small.back().time )
                {
                    temp2.time = t;
                    temp2.add = small.back().index-i;
                    ans[i].push_back(temp2);
                    small[ small.size()-1 ].time -= t;
                    t = 0;
                }
            }
            if( t > 0 )
            {
                temp1.index = i;
                temp1.time = t;
                solve.push_back(temp1);
            }
        }
    }
    for( int i = 0 ; i < solve.size() ; i++ )
    {
        int t = solve[i].time;
        for( int j = small.size()-1 ; j >= 0 ; j-- )
        {
            if( small[j].time >= t )
            {
                temp2.time = t;
                temp2.add = m-( solve[i].index-small[j].index );
                small[j].time -= t;
                ans[ solve[i].index ].push_back(temp2);
                break;
            }
            else if( t > small[j].time && small[j].time )
            {
                t -= small[j].time;
                temp2.time = small[j].time;
                small[j].time = 0;
                temp2.add = m-( solve[i].index-small[j].index );
                ans[ solve[i].index ].push_back(temp2);
            }
        }
        if( small.back().time == 0 )
            small.pop_back();
    }
    ll total = 0;
    for( int i = 1 ; i <= n ; i++ )
    {
        int len = ans[ b[i] ].size();
        if( len )
        {
            a[i] += ans[ b[i] ][len-1].add;
            total += ans[ b[i] ][len-1].add;
            ans[ b[i] ][len-1].time--;
            if( ans[ b[i] ][ len-1 ].time == 0 )
                ans[ b[i] ].pop_back();
        }
    }
    printf("%I64dn",total);
    printf("%I64d",a[1]);
    for( int i = 2 ; i <= n ; i++ )
        printf(" %I64d",a[i]);
    printf("n");
    return 0;
}
/*
10 10 3
7 1 2 2 3 3 2 1 1 7
5 4 1
1 1 2 2 3

8 4
3 3 3 1 3 3 3 1
*/