C++

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法一：

思路：

模拟

$1、通过min1,max1记录票据范围，a数组记录票据id出现次数$

$2、遍历，a[i]为0：断号，a[i]大于1：重号$

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1e5 + 10;

int n;
int a[N];
int x, min1, max1;
int x1, y1;

int main() {
    cin >> n;

    while (cin >> x) { // 不断读入直到结束
        min1 = min(min1, x);
        max1 = max(max1, x);
        a[x] ++;
    }

    for (int i = min1; i <= max1; i ++) {
        if (a[i] == 0) x1 = i;
        if (a[i] > 1)  y1 = i;
    }

    cout << x1 << ' '  << y1;

    return 0;
}

法二：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <sstream>

using namespace std;

const int N = 1e5 + 10;
int n;
int a[N];

int main() {
    int cnt;
    cin >> cnt;
    string line;

    getline(cin, line); // 忽略掉第一行的回车
    while (cnt --) {
        getline(cin, line);
        stringstream ssin(line); // 类似于 scanf("%s", )

        while (ssin >> a[n]) n ++;
    }

    sort(a, a + n);

    int res1 = 0, res2 = 0;
    for (int i = 1; i < n; i ++)
        if (a[i] == a[i - 1]) res2 = a[i]; // 重号
        else if (a[i] >= a[i - 1] + 2) res1 = a[i] - 1; // 断号

    cout << res1 << ' ' << res2;     

    return 0;
}

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