题目位置

勇敢的津津(jump)

解释代码

一句话,想说的全在代码中...

代码

//Code written by Li Zexuan (Looks handsome)
#include <bits/stdc++.h>
using namespace std;

int main() {
	freopen("jump.in", "r", stdin);
	freopen("jump.out", "w", stdout);
	int l, n, m;
	int x[500 + 1], cnt = 1;
	cin >> l >> n >> m;
	int update_m = 0; //现在跳到几米了
	for (int i = 1; i <= n; i++) {
		cin >> x[i];
	}
	x[n + 1] = l; //设置此为终点及岸边
	for (int i = 1; i <= n + 1; i++) {
		if (x[i] - update_m > m) { //如果不能一下跳过去
			cnt++; //就跳一次
			update_m = x[i - 1]; //跳到x[i-1]米了,因为x[i]跳不过去
		} else {
			continue; //否则继续直到跳不过去
		}
		
	}
	printf("%d\n", cnt);
	return 0;
}