今天來解YKL18(UVA10038)：Jolly Jumpers

Jolly Jumpers

假設輸入5 1 4 2 -1 6
代表n=5的元素，序列為[1,4,2,-1,6]
計算相鄰元素的絕對差值
|4-1|=3
|2-4|=2
|-1-2|=3
|6-(-1)|=7
檢查 1 到 n-1:
上面n-1=4
所以需要差值為{1,2,3,4}
但我們得到的差值是{3,2,7}，少了1和4，還多了7
所以=>"Not jolly"

下方我用了一個set不能重複的來表示

#include <iostream>
#include <vector>
#include <cmath>
#include <set>
using namespace std;

int main() {
    int n;
    
    while (cin >> n) {
        vector<int> seq(n);
            
        for (int i = 0; i < n; i++) {
        	//cout << i << endl;
            cin >> seq[i];
        }

        if (n == 1) {
            cout << "Jolly" << endl;
            continue;
        }
        
        //不重複
        set<int> diff;
        for (int i = 1; i < n; i++) {
            int differ = abs(seq[i] - seq[i - 1]);
            if (differ >= 1 && differ <= n - 1) {
                diff.insert(differ);
            }
        }
       
        if (diff.size() == n - 1) {
            cout << "Jolly" << endl;
        } else {
            cout << "Not jolly" << endl;
        }
    }
    return 0;
}