HDU 5427 A problem of sorting(字符串处理+排序)——BestCoder Round #54(div.2)
A problem of sorting
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≤100 which denotes the number of test cases.
For each test case, there is an positive integer n(1≤n≤100) which denotes the number of people,and next n 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 100 .Notice name only contain letter(s),digit(s) and space(s).
For each test case, there is an positive integer n(1≤n≤100) which denotes the number of people,and next n 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 100 .Notice name only contain letter(s),digit(s) and space(s).
Output
For each case, output
n lines.
Sample Input
2 1 FancyCoder 1996 2 FancyCoder 1996 xyz111 1997
Sample Output
FancyCoder xyz111 FancyCoder
Source
BestCoder Round #54 (div.2)
/************************************************************************/
附上该题对应的中文题
A problem of sorting
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
问题描述
给出一张许多人的年龄和生日表。你需要从年轻到年老输出人们的名字。（没有人年龄相同）
输入描述
第一行包含一个正整数T(T≤5)，表示数据组数。 对于每组数据，第一行包含一个正整数n(1≤n≤100)，表示人数，接下来n行，每行包含一个姓名和生日年份(1900-2015)，用一个空格隔开。姓名长度大于0且不大于100。注意姓名中只包含字母，数字和空格。
输出描述
对于每组数据，输出n行姓名。
输入样例
2 1 FancyCoder 1996 2 FancyCoder 1996 xyz111 1997
输出样例
FancyCoder xyz111 FancyCoder
/****************************************************/
出题人的解题思路：
A problem of sorting
选择任意喜欢的排序方法即可。注意名字中可能有空格。
说实话，这题还是比较简单的，初看，只要按年份从大到小排序就可以了#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 120;
const int inf = 2147483647;
const int mod = 2009;
struct person
{
char name[N];
int year;
}s[N];
bool cmp(person x,person y)
{
return x.year>y.year;
}
int main()
{
int t,n,i,j,l;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
getchar();
for(i=0;i<n;i++)
{
s[i].year=0;
gets(s[i].name);
l=strlen(s[i].name);
for(j=l-4;j<l;j++)
s[i].year=s[i].year*10+s[i].name[j]-'0';
s[i].name[l-5]='\0';
}
sort(s,s+n,cmp);
for(i=0;i<n;i++)
printf("%s\n",s[i].name);
}
return 0;
}
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