1062 talent and virtue (25 points)

1062 Talent and Virtue (25 branch )

The question

Let's start with three numbers N（ Number of members ）,L（ Pass line ）,H（ Excellent line ）
Give a set of member information , Include id, Moral character , can , Sort this group of members .
When the member's moral character and ability are not lower than H, Then the member is divided into excellent group
When the moral character of the member is not lower than H, Can not be lower than L, Then the member is divided into good groups
When the member's moral character and ability are lower than H, But no less than L And virtue is greater than ability , Then the member is divided into the medium group .
Other people are no less virtuous and talented than L, Then they are divided into two groups , Either moral character or ability is inferior to L Will not participate in the ranking .
Priority when ranking Excellent group > Good group > Medium group > Passing group
The higher the total score of morality and ability in the same group , Top of the list . The total score is the same , The higher the moral score, the more advanced . The moral character is the same id The smaller, the more forward .

Ideas

Create a member information structure , contain id, Moral character , can , The total score of morality and talent . Group all members , Sort by rules within the group .

Code

#include<stdio.h>
#include<algorithm>

using namespace std;

typedef struct node{
	int id,virtue,talent,total;
}node;

bool cmp(node x,node y)
{
	if(x.total==y.total){
		if(x.virtue==y.virtue){
			return x.id<y.id;
		}else{
			return x.virtue>y.virtue;
		}
	}else{
		return x.total>y.total;
	}
}

node sages[100007],noblemen[100007],fools[100007],others[100007];
	
int main()
{
//	node sages[100007],noblemen[100007],fools[100007],others[100007];
	int n,pass,excellent,x,y,z,count1=0,count2=0,count3=0,count4=0;
	scanf("%d%d%d",&n,&pass,&excellent);
	for(int i=0;i<n;i++){
		scanf("%d%d%d",&x,&y,&z);
		if(y>=excellent&&z>=excellent){
			sages[count1].id=x;
			sages[count1].virtue=y;
			sages[count1].talent=z;
			sages[count1].total=y+z;
			count1++;
		}else if(y>=excellent&&z>=pass){
			noblemen[count2].id=x;
			noblemen[count2].virtue=y;
			noblemen[count2].talent=z;
			noblemen[count2].total=y+z;
			count2++;
		}else if(y>=z&&z>=pass){
			fools[count3].id=x;
			fools[count3].virtue=y;
			fools[count3].talent=z;
			fools[count3].total=y+z;
			count3++;
		}else if(y>=pass&&z>=pass){
			others[count4].id=x;
			others[count4].virtue=y;
			others[count4].talent=z;
			others[count4].total=y+z;
			count4++;
		}
	}
	printf("%d\n",count1+count2+count3+count4);
	sort(sages,sages+count1,cmp);
	for(int i=0;i<count1;i++){
		printf("%d %d %d\n",sages[i].id,sages[i].virtue,sages[i].talent);
	}
	sort(noblemen,noblemen+count2,cmp);
	for(int i=0;i<count2;i++){
		printf("%d %d %d\n",noblemen[i].id,noblemen[i].virtue,noblemen[i].talent);
	}
	sort(fools,fools+count3,cmp);
	for(int i=0;i<count3;i++){
		printf("%d %d %d\n",fools[i].id,fools[i].virtue,fools[i].talent);
	}
	sort(others,others+count4,cmp);
	for(int i=0;i<count4;i++){
		printf("%d %d %d\n",others[i].id,others[i].virtue,others[i].talent);
	}
	return 0;
}