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("d %s %d\n", a[i].id, a[i].name, a[i].g); }转载于:https://www.cnblogs.com/Dup4/p/9433195.html
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