XDOJ - count the number of positive integers

完整代码如下

在这里插入#include<stdio.h>
int main()
{
    
	int n;
	scanf("%d",&n);
	int a[n];
	int i,j,temp;
	for(i=0;i<n;i++){
    
		scanf("%d",&a[i]);
	}
	for(i=0;i<n;i++){
    
		for(j=i+1;j<n;j++){
    
			if(a[i]>a[j]){
    
				temp=a[i];
				a[i]=a[j];
				a[j]=temp;
			}
		}
	}
	//The above for sorting an array,以下为核心代码 
	
  for(i=0;i<n-1;i++){
    
  	if(a[i]==a[i+1]){
    
  		for(j=i+1;j<n-1;j++){
    
  			a[j]=a[j+1];
		  }
		  i--;  //i--But the key,If once before and after a fairly,After early to judge whether to a new defender is equal.
		        //比如 
		  n--;
	  }
  }
	
	for(i=0;i<n;i++){
    
		printf("%d\n",a[i]);
	}
	return 0;
}

核心代码如下

  	if(a[i]==a[i+1]){
    
  		for(j=i+1;j<n-1;j++){
    
  			a[j]=a[j+1];
		  }
		  i--;  //i--But the key,If once before and after a fairly,After early to judge whether to a new defender is equal.
		        
		  n--;
	  }
  }

首先,This algorithm has a point to note,i在- -,同时n也必定- -;
Cases of calculus：
案例 111223
Before and after the equivalent,Ahead of time
11223,但此时i没有- -,iSmoothly into the1
显然arr[1]= arr[2]
So in the event of avant-garde is equal to the situation after a,一定要i - -
原理：From the original equal position to compare to after a,Equivalent of each has been ahead of time,Until before a is not equal to after a can from under a number as a starting point in comparison with the Numbers of the following is.
Where each comparison position as a starting point,Until when only its own a number of mobile starting point.
同时n也一定要- -,Because it must be the final 123333
Mantissa is equal,如果没有n --,3Always equal to the back of the3
,i始终- - 后又+ +
进入死循环

当然,If you feel the abstract,From the perspective of simple to understand,11223>>12233>>12333,Every do early treatment is more a do not belong to the original array's tail,The original array can not so much3,It is good to cut
补充
This is just growing up row not repeat order,The following statistical code snippet to be modified can

	for(i=0;i<n;i++){
    
		b[i]=1;
	}
  for(i=0;i<n-1;i++){
    
  	if(a[i]==a[i+1]){
    
  		for(j=i+1;j<n-1;j++){
    
  			a[j]=a[j+1];
		  }
		  b[i]++;//There have been several times the number of the first,After continuously in advance,In the starting point to the nexti才++,The second number is the second number at this time 
		  i--;  //i--But the key,If once before and after a fairly,After early to judge whether to a new defender is equal.
		        //比如 
		  n--;
	  }
  }
	
	for(i=0;i<n;i++){
    
		printf("%d:%d\n",a[i],b[i]);
	}
	return 0;
}