80Days游戏策略解析-优快云博客

本文探讨了基于儒勒·凡尔纳科幻小说《八十天环游地球》的80Days游戏策略。游戏要求玩家在有限的资金和时间内，选择最佳起点以完成环球旅行。文章分析了简化版游戏规则，包括城市间的资金获取与消耗，以及如何确保旅行全程资金充足。通过实例输入输出，展示了如何判断可能的起始城市。

时间限制:1000ms

单点时限:1000ms

内存限制:256MB

描述

80 Days is an interesting game based on Jules Verne's science fiction "Around the World in Eighty Days". In this game, you have to manage the limited money and time.

Now we simplified the game as below:

There are n cities on a circle around the world which are numbered from 1 to n by their order on the circle. When you reach the city i at the first time, you will get ai dollars (ai can even be negative), and if you want to go to the next city on the circle, you should pay bi dollars. At the beginning you have c dollars.

The goal of this game is to choose a city as start point, then go along the circle and visit all the city once, and finally return to the start point. During the trip, the money you have must be no less than zero.

Here comes a question: to complete the trip, which city will you choose to be the start city?

If there are multiple answers, please output the one with the smallest number.

输入

The first line of the input is an integer T (T ≤ 100), the number of test cases.

For each test case, the first line contains two integers n and c (1 ≤ n ≤ 106, 0 ≤ c ≤ 109). The second line contains n integers a1, …, an (-109 ≤ ai ≤ 109), and the third line contains n integers b1, …, bn (0 ≤ bi ≤ 109).

It's guaranteed that the sum of n of all test cases is less than 106

输出

For each test case, output the start city you should choose.

提示

For test case 1, both city 2 and 3 could be chosen as start point, 2 has smaller number. But if you start at city 1, you can't go anywhere.

For test case 2, start from which city seems doesn't matter, you just don't have enough money to complete a trip.

样例输入

3 0

3 4 5

5 4 3

3 100

-3 -4 -5

30 40 50

样例输出

-1

#include<iostream>
using namespace std;
long long a[2000002],b[2000002],m[2000002];
int main(){
	int t;
	cin>>t;
	while(t--){
		long n,c;
		cin>>n>>c;
		for(int i=1;i<=n;i++)
			cin>>a[i];
		for(int i=1;i<=n;i++)
			a[n+i]=a[i];	
		for(int j=1;j<=n;j++)
			cin>>b[j];
		for(int j=1;j<=n;j++)
			b[j+n]=b[j];
		for(long i=1;i<=2*n;i++){
			m[i]=a[i]-b[i];
		}
		int g=0;
		for(long i=1;i<=n;i++){
			long r=c;
			bool f=true;
			for(long s=i;s<i+n;s++){
				r=r+m[s];
				//r=r+a[s];
				//r=r-b[s];
				if(r<0){
					f=false;
					break;	
				}
			}
			if(f==true){
				cout<<i<<endl;
				g=1;
				break;
			}
		}
		if(g==0)
			cout<<"-1"<<endl;
	}
	
}