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ZOJ - 4104 sequence in the pocket

2022-06-24 16:00:00 【51CTO】

Original link ： https://zoj.pintia.cn/problem-sets/91827364500/problems/91827370499

ZOJ——4104 Sequence in the Pocket（ Thinking problems ）_ios
The test sample ：

Sample Input
2
4
1 3 2 4
5
2 3 3 5 5
Sample Output
2
0

Sample explanation ：

For the first sample test case , Move the third element to the front （ So the sequence becomes {2,1,3,4}）, Then move the second element to the front （ So the sequence becomes {1, 2, 3,4}）. Now? , The sequence is not decreasing .
For the second sample test case , Since the sequence has been sorted , Therefore, no operation is required .

The question ： Here's a sequence of integers , You can move an element in the sequence to the front of the sequence any number of times , Ask how many times you have to go through at least to make this sequence not decreasing .

Their thinking ： The topic is really simple , Our goal is to make the sequence not decreasing , So what if you encounter a decreasing ？ Should we arrange it in order so that the sequence does not decrease , But not everyone has to operate ？ That must not be , Our final sequence state is our ordered sequence , Then we use this ordered sequence to compare with the original sequence from back to front （ The comparison from back to front is because our operation is to put in front of the sequence ）, Just judge which ones are not operated , So the other thing is to make the sequence non decreasing by moving to the front of the sequence .OK, Let's look at the code .

AC Code :

       
       /*
       
       *
       
       */
       
       #include<bits/stdc++.h> 
       
       //POJ I won't support it 
       
       #define rep(i,a,n) for (int i=a;i<=n;i++)
       
       //i Is a cyclic variable ,a For the initial value ,n Is the limit value , Increasing 
       
       #define per(i,a,n) for (int i=a;i>=n;i--)
       
       //i Is a cyclic variable , a For the initial value ,n Is the limit value , Decline .
       
       #define pb push_back
       
       #define IOS ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
       
       #define fi first
       
       #define se second
       
       #define mp make_pair
       
       using 
       
       namespace 
       
       std;
       
       const 
       
       int 
       
       inf 
       
       = 
       
       0x3f3f3f3f;
       
       // infinity 
       
       const 
       
       int 
       
       maxn 
       
       = 
       
       1e5;
       
       // Maximum .
       
       typedef 
       
       long 
       
       long 
       
       ll;
       
       typedef 
       
       long 
       
       double 
       
       ld;
       
       typedef 
       
       pair
       
       <
       
       ll, 
       
       ll
       
       >  
       
       pll;
       
       typedef 
       
       pair
       
       <
       
       int, 
       
       int
       
       > 
       
       pii;
       
       //******************************* Split line , The above is a custom code template ***************************************//
       
       int 
       
       t,
       
       n,
       
       a[
       
       maxn],
       
       b[
       
       maxn];
       
       int 
       
       main(){
       
       //freopen("in.txt", "r", stdin);// When submitting, you should comment out 
       
       IOS;
       
       while(
       
       cin
       
       >>
       
       t){
       
       while(
       
       t
       
       --){
       
       cin
       
       >>
       
       n;
       
       rep(
       
       i,
       
       0,
       
       n
       
       -
       
       1){
       
       cin
       
       >>
       
       a[
       
       i];
       
       b[
       
       i]
       
       =
       
       a[
       
       i];
       
      }
       
       sort(
       
       b,
       
       b
       
       +
       
       n);
       
       // Sort 
       
       int 
       
       pos
       
       =
       
       n
       
       -
       
       1,
       
       sum
       
       =
       
       0;
       
       //sum Count the number of times we have to move to the front 
       
       per(
       
       i,
       
       n
       
       -
       
       1,
       
       0){
       
       if(
       
       a[
       
       i]
       
       ==
       
       b[
       
       pos])
       
       pos
       
       --;
       
       // There is no need to move forward 
       
       else  
       
       sum
       
       ++;
       
       // Operation forward 
       
      }
       
       cout
       
       <<
       
       sum
       
       <<
       
       endl;
       
    }
       
  }
       
       return 
       
       0;
       
}
      
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