BestCoder Round #54 (div.2) HDOJ 5427 A problem of sorting(模拟)

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

21FancyCoder 19962FancyCoder 1996xyz111 1997

 

Sample Output

FancyCoderxyz111FancyCoder

 

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;}