HDU 5427 A problem of sorting 水题

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 }

View Code