1062 Talent and Virtue (25 分)
Ideas
The classical Chinese text of this question showed me confused. The
core of the sort is sort, which needs to be grouped in advance. I opened four arrays for the four types. In fact, you can add a flag to the struct to mark different groups for classification. The code can be more concise.
Code
#include<bits/stdc++.h>
using namespace std;
struct stu{
int id;
int vir,tal;
int sum;
}good1[100005],good2[100005],good3[100005],good4[100005];
bool cmp(stu a,stu b)//int为数组数据类型
{
if(a.sum !=b.sum)return a.sum>b.sum;//降序排列
else if(a.vir !=b.vir)return a.vir>b.vir;
else return a.id<b.id;
//return a<b;//默认的升序排列
}
//输入先分好类 再调用sort进行排序
int main()
{
int id,n,grade1,grade2;
cin>>n>>grade1>>grade2;
int n1=0,n2=0,n3=0,n4=0;
int g1,g2;
int abc = n;
for(int i=0;i<n;i++)
{
cin>>id>>g1>>g2;
if(g1<grade1 || g2<grade1)
{
abc--;
continue;
}
else if(g1>=grade2 && g2>=grade2)
{
good1[n1].id = id;
good1[n1].vir = g1;
good1[n1].tal = g2;
good1[n1++].sum = g1 +g2;
}
else if(g1>=grade2 && g2<grade2)
{
good2[n2].id = id;
good2[n2].vir = g1;
good2[n2].tal = g2;
good2[n2++].sum = g1 +g2;
}
else if(g1<grade2 && g1>=g2 && g2<grade2)
{
good3[n3].id = id;
good3[n3].vir = g1;
good3[n3].tal = g2;
good3[n3++].sum = g1 +g2;
}
else
{
good4[n4].id = id;
good4[n4].vir = g1;
good4[n4].tal = g2;
good4[n4++].sum = g1 +g2;
}
}
sort(good1,good1+n1,cmp);
sort(good2,good2+n2,cmp);
sort(good3,good3+n3,cmp);
sort(good4,good4+n4,cmp);
cout<<abc<<endl;
for(int i =0;i<n1;i++)
cout<<good1[i].id<<" "<<good1[i].vir<<" "<<good1[i].tal<<endl;
for(int i =0;i<n2;i++)
cout<<good2[i].id<<" "<<good2[i].vir<<" "<<good2[i].tal<<endl;
for(int i =0;i<n3;i++)
cout<<good3[i].id<<" "<<good3[i].vir<<" "<<good3[i].tal<<endl;
for(int i =0;i<n4;i++)
cout<<good4[i].id<<" "<<good4[i].vir<<" "<<good4[i].tal<<endl;
}