2020 BIT冬训-C++STL C - A problem of sorting HDU - 5427

 

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 nn 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 nn lines.

 

Sample Input

2
1
FancyCoder 1996
2
FancyCoder 1996
xyz111 1997

Sample Output

FancyCoder
xyz111
FancyCoder

这题的坑点是名字也有空格！这里需要掌握的知识是字符串的截取，以及需要注意在字符串后面添‘\0’

#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
struct people{
    char name[105];
    int birth;
}p[105];
int cmp(people a,people b){
    if(a.birth == b.birth)
        return strcmp(a.name,b.name)<0;
    else
        return a.birth>b.birth;
}
char data[115];
int t,n;
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        cin.get();
        for(int i=0;i<n;i++){
            cin.getline(data,1024);
            p[i].birth=0;
            for(int k = strlen(data) - 4; k < strlen(data); k++){
                p[i].birth *= 10;
                p[i].birth += data[k] - '0';
            }
            data[strlen(data)-5]='\0';
            strcpy(p[i].name,data);
        }
        sort(p,p+n,cmp);
        for(int i=0;i<n;i++)
            printf("%s\n",p[i].name);
    }
    return 0;
}