Pat 1055 first k

The second group of data is relatively large. If it is simply sorted, direct retrieval times out because all data is traversed each time.

N/200 = 500, indicating that up to 500 people of the same age can be found, while M = 100 is relatively small, meaning that all people of the same age after 100 will not be found.

 

#include<iostream>#include<cstring>#include<cstdio>#include <algorithm>using namespace std;struct node{char name[10];int age,worth;}a[100005];int b[20005];int c[205];bool cmp2(const node& p,const node& q){if(p.worth==q.worth){if (p.age==q.age){return strcmp(p.name,q.name)<=0;}return p.age<q.age;}return p.worth>q.worth;}int main(){int n,q,i,M,ma,mb,txt=1,len;while(~scanf("%d %d",&n,&q)){memset(c,0,sizeof(c));for (i=0;i<n;++i){scanf("%s%d%d",a[i].name,&a[i].age,&a[i].worth);}sort(a,a+n,cmp2);for (len=i=0;i<n;++i){if (c[a[i].age]<=100){b[len++]=i;c[a[i].age]++;}}// printf("----\n");// for (i=0;i<n;++i){// printf("%s %d %d\n",a[b[i]].name,a[b[i]].age,a[b[i]].worth);// }while(q--){scanf("%d%d%d",&M,&ma,&mb);if(ma>mb){ma^=mb;mb^=ma;ma^=mb;}printf("Case #%d:\n",txt++);int cnt=0;for (i=0;i<len&&cnt<M;++i){if(a[b[i]].age>=ma&&a[b[i]].age<=mb){printf("%s %d %d\n",a[b[i]].name,a[b[i]].age,a[b[i]].worth);cnt++;}}if(cnt==0){printf("None\n");}}}return 0;}