题目链接

https://www.patest.cn/contests/pat-a-practise/1028

思路

就按照 它的三种方式 设计 comp 函数 然后快排就好了

但是 如果用 c++ 中的 string 保存名字的话 就会超时

所以 用 c 里面的 char *s 就可以过

AC代码

#include <cstdio>

#include <cstring>

#include <ctype.h>

#include <cstdlib>

#include <cmath>

#include <climits>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <deque>

#include <vector>

#include <queue>

#include <string>

#include <map>

#include <stack>

#include <set>

#include <numeric>

#include <sstream>

#include <iomanip>

#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))

#define pb push_back

using namespace std;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;

typedef pair <int, int> pii;

typedef pair <ll, ll> pll;

typedef pair<string, int> psi;

typedef pair<string, string> pss;

const double PI = 3.14159265358979323846264338327;

const double E = exp(1);

const double eps = 1e-30;

const int INF = 0x3f3f3f3f;

const int maxn = 1e5 + 5;

const int MOD = 1e9 + 7;

struct Node

{

    int id, g;

    char name[10];

}a[maxn];

bool comp_1(Node x, Node y)

{

    return x.id < y.id;

}

bool comp_2(Node x, Node y)

{

    if (strcmp(x.name, y.name) == 0)

        return x.id < y.id;

    return strcmp(x.name, y.name) > 0 ? 0 : 1;

}

bool comp_3(Node x, Node y)

{

    if (x.g == y.g)

        return x.id < y.id;

    return x.g < y.g;

}

int main()

{

    int n, c;

    scanf("%d%d", &n, &c);

    for (int i = 0; i < n; i++)

        scanf("%d%s%d", &a[i].id, &a[i].name, &a[i].g);

    if (c == 1)

        sort(a, a + n, comp_1);

    else if (c == 2)

        sort(a, a + n, comp_2);

    else if (c == 3)

        sort(a, a + n, comp_3);

    for (int i = 0; i < n; i++)

        printf("%06d %s %d\n", a[i].id, a[i].name, a[i].g);

}