题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5427
A problem of sorting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1447 Accepted Submission(s): 554
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).
Output
For each case, output n lines.
Sample Input
2
1
FancyCoder 1996
2
FancyCoder 1996
xyz111 1997
Sample Output
FancyCoder
xyz111
FancyCoder
题解：
被水题pe了一早上，学了个fgets来取代gets，据说后者不太安全，最好不要用。
代码：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 using namespace std;
6
7 const int maxn = 111;
8
9 char name[maxn][maxn];
10 int age[maxn];
11 int ran[maxn];
12
13 int n;
14
15 bool cmp(int x, int y) {
16 return age[x] > age[y];
17 }
18
19 void init() {
20
21 }
22
23 int main() {
24 int tc;
25 scanf("%d", &tc);
26 while (tc--) {
27 //用scanf("%d\n",&n);代替以下两行会PE。
28 scanf("%d", &n);
29 getchar();
30 for (int i = 0; i < n; i++) {
31 //fgets得到的字符串，在‘\0‘之前会多一位‘\n‘!
32 fgets(name[i], sizeof(name[i]), stdin);
33 // gets(name[i]); 不安全，不要用
34 int len = strlen(name[i]);
35 age[i] = 0;
36 for (int j = len - 5; j < len-1; j++) {
37 age[i] = age[i] * 10 + name[i][j] - ‘0‘;
38 }
39 name[i][len - 6] = ‘\0‘;
40 }
41 for (int i = 0; i < n; i++) {
42 //printf("str:%s.age:%d\n", name[i],age[i]);
43 }
44 for (int i = 0; i < n; i++) ran[i] = i;
45 sort(ran, ran + n, cmp);
46 for (int i = 0; i < n; i++) {
47 int x = ran[i];
48 printf("%s\n", name[x]);
49 }
50 }
51 return 0;
52 }