Best code #54 div 2 A water

A Problem of sortingaccepts:443submissions:1696Time limit:2000/1000 MS (java/others)Memory limit:65536/65536 K (java/others)Problem Description

There is 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 both have the the same age.)

Input

First line contains a single integer t≤100t \leqT≤100 which denotes the number of test CA Ses.

For each test case, there was an positive integern (1≤n≤100) n (1 \leq n \leq) n (1≤ n≤100< Span class= "Mclose") which denotes the number of People,and next nn< Span class= "katex-html" >n Lines,each Line have 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

21FancyCoder 19962FancyCoder 1996xyz111 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;}

Best code #54 div 2 A water