A problem of sorting
Accepts: 443
Submissions: 1696
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
Problem Description
There are many people's name and birth in a list.Your task is to print the name from young to old.(There is no pair of two has the same age.)
Input
First line contains a single integer T≤100T \leq 100T≤100 which denotes the number of test cases.
For each test case, there is an positive integer n(1≤n≤100)n (1 \leq n \leq 100)n(1≤n≤100) which denotes the number of people,and next nnn lines,each line has a name and a birth's year(1900-2015) separated by one space.
The length of name is positive and not larger than 100100100.Notice name only contain letter(s),digit(s) and space(s).
Output
For each case, output nnn lines.
Sample Input
2 1 FancyCoder 1996 2 FancyCoder 1996 xyz111 1997
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<map> #include<queue> using namespace std; struct node { char name[105]; int birthday; friend bool operator <(node a,node b) { return a.birthday<b.birthday; } }; node q1,q2; priority_queue<node>q; int main() { int t,n; while(scanf("%d",&t)!=EOF) { for(int i=1; i<=t; i++) { scanf("%d",&n); getchar(); for(int j=1; j<=n; j++) { char str[105],strr[105]; int birth; memset(strr,0,sizeof(strr)); memset(str,0,sizeof(str)); cin.getline(str,105); int len=strlen(str); birth=str[len-1]-'0'+(str[len-2]-'0')*10+(str[len-3]-'0')*100+(str[len-4]-'0')*1000; for(int k=0; k<=len-1-5; k++) { strr[k]=str[k]; } strcpy(q1.name,strr); q1.birthday=birth; q.push(q1); } while(!q.empty()) { q2=q.top(); q.pop(); cout<<q2.name<<endl; } } } return 0; }