BestCoder Round #54 (div.2) HDOJ 5427 A problem of sorting(模拟)_a problem of sorting bestcoder round #54 (div.2)-优快云博客

本文讨论了一个关于排序的问题，包括时间限制、内存限制等参数。文章详细解释了如何通过输入和输出来解决这个问题，涉及年龄排序和名字排序的实现过程。通过简单的模拟，解决了排序问题，并提供了AC代码作为示例。

A problem of sorting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 240 Accepted Submission(s): 115

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

简单的模拟，注意名字中含有空格，gets()函数就搞定，还有pow()函数原型也要注意一下。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "cmath"
using namespace std;
const int MAXN = 105;
struct node
{
	/* data */
	char name[200];
	int age;
}a[MAXN];
bool cmp(node a, node b)
{
	return a.age > b.age;
}
int main(int argc, char const *argv[])
{
	int t;
	scanf("%d", &t);
	for(int cas = 0; cas < t; ++cas) {
		memset(a, 0, sizeof(a));
		int n;
		cin >> n;
		getchar();
		for(int i = 0; i < n; ++i) {
			gets(a[i].name);
			int len = strlen(a[i].name), num = 0;
			for(int j = len - 1; j >= len - 4; --j)
				a[i].age += ((int)(pow(10.0, num++))) * (a[i].name[j] - '0');
		}
		sort(a, a + n, cmp);
		for(int i = 0; i < n; ++i) {
			int len = strlen(a[i].name);
			a[i].name[len - 5] = '\0';
			printf("%s\n", a[i].name);
		}
	}
	return 0;
}