hdu 5427 A problem of sorting

2023-08-27 17:38:34

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5427

A problem of sorting

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 $T \leq 100$ which denotes the number of test cases.

For each test case, there is an positive integer $n\ (1 \leq n \leq 100)$ 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

字符串简单题。。

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<vector>

#include<queue>

#include<set>

using std::set;

using std::sort;

using std::pair;

using std::swap;

using std::queue;

using std::multiset;

#define pb(e) push_back(e)

#define sz(c) (int)(c).size()

#define mp(a, b) make_pair(a, b)

#define all(c) (c).begin(), (c).end()

#define iter(c) decltype((c).begin())

#define cls(arr, val) memset(arr, val, sizeof(arr))

#define cpresent(c, e) (find(all(c), (e)) != (c).end())

#define rep(i, n) for(int i = 0; i < (int)n; i++)

#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)

const int N = 110;

const int INF = 0x3f3f3f3f;

typedef unsigned long long ull;

struct Node {

	int age;

	char name[N];

	inline bool operator<(const Node &x) const {

		return age > x.age;

	}

}A[N];

int main() {

#ifdef LOCAL

	freopen("in.txt", "r", stdin);

	freopen("out.txt", "w+", stdout);

#endif

	int t, n;

	char buf[200], digit[5];

	scanf("%d", &t);

	while (t--) {

		scanf("%d", &n);

		getchar();

		rep(i, n) {

			gets(buf);

			int v, j = 0, len = strlen(buf);

			for (; j < 4; j++) digit[j] = buf[len - 4 + j];

			digit[j] = '\0'; v = atoi(digit);

			buf[len - 5] = '\0';

			A[i].age = v, strcpy(A[i].name, buf);

		}

		sort(A, A + n);

		rep(i, n) printf("%s\n", A[i].name);

	}

	return 0;

}

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