例题8-13(uva-11093)

最新推荐文章于 2021-05-25 13:48:21 发布

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最新推荐文章于 2021-05-25 13:48:21 发布

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分类专栏：算法入门第八章

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本文介绍了一个使用C++解决的问题：通过编程确定燃料补给站能够完成环形路线行驶的最晚出发点。此问题涉及数组操作、条件判断及简单的数学运算。

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#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int maxn = 1e5 + 7;
int n,input[maxn],req[maxn];

int main()
{
	int t, kcase = 0;
	cin >> t;
	while (t--) {
		cin >> n;
		long long int sum = 0,nMaxReq = 0,ans = 1;
		for (int i = 1; i <= n; i++) {
			cin >> input[i];
		}
		for (int i = 1;i<=n;i++){
			cin >> req[i];
		}
		for (int i = 1;i<=n;i++)
		{
			sum += input[i];
			if (sum < req[i]) {
				nMaxReq += req[i] - sum;
				ans = i + 1;
				sum = 0;
			}
			else sum -= req[i];
		}
		cout << "Case " << ++kcase << ": ";
		if (ans == n + 1 || sum < nMaxReq)
			cout << "Not possible" << endl;
		else
			cout << "Possible from station " << ans << endl;
	}
	return 0;
}