difficult

codeup

To Fill or Not to Fill

初始化每个加油站的start和end，遍历每个加油站可到达的加油站，比较价格，修改start和end。

只过了pat

//codeup To Fill or Not to Fill
//汽油容量 距离 每个gas可跑距离 汽油站数量
//59	  525	 19				2
//单价	   距离
//3			314
//3			 0
#include<iostream>
#include<algorithm>
using namespace std;
typedef struct station {
	double price;
	double start;
	double end;
};
bool cmp(station a, station b) {
	return a.start < b.start;
}
int main() {
	int max, meter, per_meter, n;
	station sta[510];
	while (cin >> max >> meter >> per_meter >> n) {
		for (int i = 0; i < n; i++) {
			cin >> sta[i].price >> sta[i].start;
			sta[i].end = sta[i].start + max * per_meter;
		}

		sort(sta, sta + n, cmp);
		/*for (int i = 0; i < n; i++) {
			cout << "排序结果\n";
			cout << sta[i].start << "  " << sta[i].end << " "<<sta[i].price<<endl;
		}*/
		if (sta[0].start != 0 || n == 0) {
			cout << "The maximum travel distance = ";
			printf("%.2f\n",0.00);
			break;
		}
		for (int i = 0; i < n; i++) {
			if (sta[i].end > meter) {
				sta[i].end = meter;
			}
			for (int j = i + 1; j < n; j++) {
				if (sta[j].start < sta[i].end) {
					if (sta[j].price <= sta[i].price) {//新的更便宜
						if(sta[i].start>sta[j].start &&sta[i].end<sta[j].end){//被包围
							sta[i].end = sta[i].start;
							break;
						}
						sta[i].end = sta[j].start;
					}
					else {
						sta[j].start = sta[i].end;
					}
				}
				else {//没有加油站了
					break;
				}
			}
		}
		double money = 0, all_meter = 0;
		for (int i = 0; i < n; i++) {
			//cout << sta[i].start << "  " << sta[i].end << endl;
			all_meter += sta[i].end - sta[i].start;
			money += (sta[i].end - sta[i].start) / per_meter * sta[i].price;
		}
		if (all_meter < meter) {
			cout << "The maximum travel distance = ";
			printf("%.2f\n", all_meter);
		}
		else {
			printf("%.2f\n", money);
		}
	}
}

---

【递归入门】出栈序列统计

是一道深度优先的递归题，抄了答案。
[8.1小节递归入门E]: http://codeup.hustoj.com/problem.php?cid=100000608&pid=4

#include <iostream>
using namespace std;
int n;
int cnt;
void f(int in, int out) {
	if (out > n) {
		cnt++;
		return;
	}
	if (in > n || out > in) {
		return;
	}
	else {
		f(in + 1, out);
		f(in, out + 1);
	}
}
int main() {
	while (cin >> n) {
		cnt = 0;
		f(0, 0);
		cout << cnt << endl;
	}
}