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：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn = 111;

char name[maxn][maxn];
int age[maxn];
int ran[maxn];

int n;

bool cmp(int x, int y) {
    return age[x] > age[y];
}

void init() {

}

int main() {
    int tc;
    scanf("%d", &tc);
    while (tc--) {
        //用scanf("%d\n",&n);代替以下两行会PE。 
        scanf("%d", &n);
        getchar();
        for (int i = 0; i < n; i++) {
            //fgets得到的字符串，在'\0'之前会多一位'\n'! 
            fgets(name[i], sizeof(name[i]), stdin);
//            gets(name[i]); 不安全，不要用 
            int len = strlen(name[i]);
            age[i] = 0;
            for (int j = len - 5; j < len-1; j++) {
                age[i] = age[i] * 10 + name[i][j] - '0';
            }
            name[i][len - 6] = '\0';
        }
        for (int i = 0; i < n; i++) {
            //printf("str:%s.age:%d\n", name[i],age[i]);
        }
        for (int i = 0; i < n; i++) ran[i] = i;
        sort(ran, ran + n, cmp);
        for (int i = 0; i < n; i++) {
            int x = ran[i];
            printf("%s\n", name[x]);
        }
    }
    return 0;
}