Codeforces Round # 584 (div.1 + div.2) (Bu Ti)

CF problem solution blog: https://codeforces.com/blog/entry/69791?tdsourcetag=s_pcqq_aiomsg

Replenishment of ...

A. Paint the Numbers

Meaning of the questions:

To a sequence, wherein a selected few number satisfying a divisor of the entire sequence can be selected. Numbers by the selected minimum number.

Ideas:

Violence enumerate all combinations of elements (title in value is relatively small).

code:

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
const int mod = 1e9+7;
const int maxn = 108;
 
int a[maxn];
int cnt;
bool vis[maxn];

bitset<101> bs;
 
int main(){
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++){
        scanf("%d",a+i);
    }
    sort(a+1,a+1+n);
    int ans=0;
    int flag=0;
    bs.reset();
    for(int i=1; i<=n; i++){
        flag=0;
        for(int j=i; j<=n; j++){
            if(a[j]%a[i]==0 && !bs[j]){
                bs[j] = 1;
                flag = 1;
            }
        }
        if(flag){
            ans++;
        }
        if(bs.count()==n){
            printf("%d\n", ans);
            return 0;
        }
    }
    printf("%d\n", ans);
}

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B. Koala and Lights

Meaning of the questions:

N you lamp in the initial state (on or off), and for each lamp is given two values ​​a, b; a representative of fast off state of the switching cycle, b represents the first switching time. Q. At some point, the number of bulbs lit simultaneously, and up to how much.

Ideas:

Violence: 1e5 enumeration of all time, then the statistical maximum.

code:

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
const int mod = 1e9+7;
const int maxn = 108;
 
char s[maxn]; 
int a[maxn],b[maxn],c[maxn];
bitset<maxn> bs;
 
int main(){
    int n;
    scanf("%d",&n);
    scanf("%s",s+1);
    bs.reset();
    for(int i=1; i<=n; i++){
        bs[i] = s[i]-'0';
    }
    for(int i=1; i<=n; i++){
        scanf("%d%d",a+i,b+i);
    }
    int ans=0;
    for(int j=0; j<=30000; j++){
        for(int i=1; i<=n; i++){
            if( j>=b[i] && ((j-b[i])%a[i] )==0 )  {
                bs[i] = (int)bs[i]^1;
            }
        }
        ans = max(ans,(int)bs.count());
    }
    printf("%d\n", ans);
}

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C. Paint the Digits

Meaning of the questions:

A given sequence, then select two sequences, two sequences to ensure itself (non-decreasing), and finally to ensure that the selected 12 sequence also showed a (non-decreasing) arrayed

Ideas:

A non-descending order, sequence. It is easy to think of monotonous stack. So we start and monotonous side queue to find out all eligible 1. 1 but also to the queue for processing, i.e., to find the first element 2, is greater than all the queues of all of the first element 2 out of the stack,

This would address 12 is also in a non-descending order. For the case where no solution that is determined for the first time after the handle 1 if the sequence remaining in a non-decreasing can.

code:

#include <cstdio>
#include <queue>
#include <iostream>
#include <algorithm>
#include <stack>
#include <string>
#include <cstring> 
using namespace std;
const int maxn  =2e5+7;
const int inf = 0x3f3f3f3f;
char arr[maxn];
stack<int>a;
int top;
struct node{
    int val;
    int cur;
}s[maxn]; 
 
void init(){
    top = 0;
    s[0].val = -1; s[0].cur = -1;
    while(!a.empty()) a.pop();        
} 
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d",&n);
        scanf("%s",arr); 
        init();
        for(int i=0;i<n;i++){
            while(s[top].val>arr[i]-'0'){
                top--;
            }
            s[++top].val = arr[i]-'0';
            s[top].cur = i;
        }
        int first=-1;//最后最小位置 
        int cu=1;
        for(int i =0;i<n;i++){
            if(s[cu].cur==i) cu++;
            else{
                first = i;
                break;
            } 
        }
        while(s[top].val>arr[first]-'0'&&first!=-1){
            top--;
        }
        cu = 1;
        for(int i =0;i<n;i++){
            if(s[cu].cur==i&&cu<=top) {
                //找到第一个最小位置 
                cu++;
            }else a.push(arr[i]-'0');
        }
        int flag = 0; 
        if(!a.empty()){
            int now = a.top(); a.pop();
            int len = a.size();
            for(int i=0;i<len;i++){
                int tmp = a.top();
                a.pop();
                if(now<tmp) {
                    flag = 1; break;
                }
                now = tmp;
            }
        }
        if(flag){
            puts("-");
        }else{
            cu = 1;
            for(int i =0;i<n;i++){
                if(i==s[cu].cur&&cu<=top){
                    printf("1");
                    cu++;
                }else printf("2");
            }
            puts("");
        }
    }
}

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