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＃797div3 A---C

2022-07-08 02:04:00 【51CTO】

A

#797div3 A---C_i++

Code ：

       
       /*
       
 It's the champion , The bronze , The platform of the runner up . Middle high , Lower the left , The lowest on the right 
       
1 The platform must be higher than 2 platform ,2 Must be higher than 3
       
3 Both platforms should be larger than 0（ At least one 
       
 Purpose ： Give total n, Ask the middle one to be as small as possible . Output any one 
       
*/
       
#include <bits/stdc++.h>
       
using namespace std;
       
const int maxn = 1e9 + 5;
       
void s(){
       
  int n;
       
  cin >> n;
       
  int a,b,c;
       
  for(int i = 1; i <= n ;i ++){
       
    if(i-1>0&&i>0&&n-2*i+1>0){// structure a = i -1; b = i; c = n - a - b
       
      a = i - 1;
       
      b = i;
       
      c = n-2*i+1;  
       
      if(a > c && c > 0){
       
        break;
       
      }       
       
    }
       
  }
       
  for(int i = 1; i <= n; i++){
       
    if(a-1>c+2){// Adjust here c, Give Way c Try to rise up 
       
      a-=1, b-=1;
       
      c+=2;     
       
    }else if(a == c){// A special judge 2 3 2; Output is 2 4 1 Examples 
       
      c-=1,b+=1;
       
    }
       
  }
       
  cout <<a << " " << b << " " << c << '\n';
       
}
       
int main(){
       
  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
       
  #ifdef LOCAL
       
  FILE *p = fopen("input.txt", "a+");
       
  fclose(p);
       
  freopen("input.txt", "r", stdin);
       
//  freopen("output.txt", "w", stdout);
       
  #endif
       
  int t; cin >> t;
       
  while(t--) s();
       
  return 0;
       
}
       
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B

#797div3 A---C_i++_02

Code ：

       
       #define LOCAL
       
/*
       
 Two positive arrays a,b
       
 It can be operated countless times a Array ：
       
 Every non-zero element -1
       
 Purpose ： Determine whether you can get an array b, Make every element equal 
       
*/
       
#include <bits/stdc++.h>
       
using namespace std;
       
const int maxn = 5e4 + 5, INF = 1e9 + 5;
       
int data[maxn], data2[maxn];
       
void s(){
       
  int n ; cin >> n;
       
  int maxone = -INF, sp = 0;
       
  for(int i = 1;i <= n;i ++){
       
    cin >> data[i];
       
    if(data[i] > maxone) maxone = data[i], sp = i;// Find a maximum 
       
  }
       
  for(int i = 1; i <= n; i++) cin >> data2[i];
       
  int chazhi = maxone - data2[sp];
       
  if(chazhi < 0) {// If the difference is negative , Just no
       
    cout << "NO\n";
       
    return;
       
  }
       
  for(int i = 1; i <= n ; i ++){
       
    int tmp = (data[i] - chazhi >= 0 ? data[i] - chazhi : 0 );// You can merge and write , Just judge whether 0 Or make the difference equal 
       
    if(data2[i] != tmp) {
       
      cout << "NO\n";
       
      return;
       
    }
       
  }
       
  cout << "YES\n";
       
}
       
int main(){
       
  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
       
  #ifdef LOCAL
       
  FILE *p = fopen("input.txt", "a+");
       
  fclose(p);
       
  freopen("input.txt", "r", stdin);
       
//  freopen("output.txt", "w", stdout);
       
  #endif
       
  int t; cin >> t;
       
  while(t--) s();
       
  return 0;
       
}
       
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C

#797div3 A---C_i++_03

Code ：

       
       #define LOCAL
       
/*
       
 complete n A mission ：
       
 The first i Time s Yes, I know. . No two tasks are at the same time 
       
 The first i personal f, It is the time point when the task is completed 
       
 I don't know the duration of the task 
       
 It is said that , Tasks are completed on a first come, first served basis 
       
 Once you come to the first task , It's going to be implemented 
       
 If a task arrives when the previous task has not been completed , Just put the new task at the end of the queue and wait 
       
 If he finishes , You are about to perform the next task and the queue is not empty , He immediately took the team leader element . Wait before the task comes 
       
 Purpose ： Find out how long each task has been performed 
       
*/
       
#include <bits/stdc++.h>
       
using namespace std;
       
const int maxn = 2e5 + 5;
       
struct per{
       
  int s, f;
       
} data[maxn];
       
void s(){
       
  int n ; cin >> n;
       
  for(int i = 1; i <= n; i++) cin >> data[i].s;
       
  for(int i = 1; i <= n; i++) cin >> data[i].f;
       
  int nowtime = 0;
       
  for(int i = 1 ; i <= n; i++){// Simulate a task processing process 
       
    if(nowtime < data[i].s) nowtime = data[i].s;// The current time can jump , If you have to wait , Then jump directly to the beginning 
       
    if(data[i].f - nowtime > 0){// If there is still a long life span for this task , Just output the answer 
       
      cout << data[i].f - nowtime <<" ";
       
    }
       
    else if(data[i].f - nowtime <= 0){// If this task is out of date , It outputs 0
       
      cout << 0 << " ";
       
    }
       
    nowtime = data[i].f;// Update the current time to finish the current task 
       
  }
       
  cout <<'\n';
       
}
       
int main(){
       
  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
       
  #ifdef LOCAL
       
  FILE *p = fopen("input.txt", "a+");
       
  fclose(p);
       
  freopen("input.txt", "r", stdin);
       
//  freopen("output.txt", "w", stdout);
       
  #endif
       
  int t; 
       
  cin >> t;
       
  while(t--) s();
       
  return 0;
       
}
       
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