題目連結：http://hihocoder.com/problemset/problem/1831

時間限制:1000ms

單點時限:1000ms

記憶體限制: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.

輸入

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

輸出

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.

樣例輸入

2
3 0
3 4 5
5 4 3
3 100
-3 -4 -5
30 40 50

樣例輸出

2
-1

EmacsNormalVim

題目大意：T組資料，每組資料n，c（初始金幣數量），n個城市（編號1-n）按輸入順序圍成一個環，每個城市進入獲得ai金幣，出城市失去bi金幣（都可能是負數）問從哪個城市開始走，能保證走完這個環時不出現負的金幣持有量，多個答案取最小的那個編號（如果沒有符合要求的，輸出-1）：

很容易就想到把一個環展開，1-n-2*n，取其中的n連續的n個數就是一個環；

然後對於每個點，如果符合要求，則新增進佇列，如果不符合要求，則取出隊首元素，再次檢視；

程式碼比較簡單：

//#pragma comment(linker, "/STACK:1024000000,1024000000") 
 
#include<stdio.h>
#include<string.h>  
#include<math.h>  
#include<stdlib.h>

//#include<map>   
//#include<set>
#include<deque>  
#include<queue>  
#include<stack>  
#include<bitset> 
#include<string>  
#include<fstream>
#include<iostream>  
#include<algorithm>  
using namespace std;  
 
#define ll long long  
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b) 
#define clean(a,b) memset(a,b,sizeof(a))// 水印 
//std::ios::sync_with_stdio(false);
const int MAXN=1e6+10;
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;

ll arr1[MAXN<<1],arr2[MAXN<<1];
deque<int> que;
//void intt()
//{
//	clean(arr1,0);
//	clean(arr2,0);
//}
int main()
{
	std::ios::sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--)
    {
    	//intt();
    	int n;
    	ll sum;
    	cin>>n>>sum;
    	for(int i=1;i<=n;++i)//輸入in 
    		cin>>arr1[i];
    	for(int i=n+1;i<=n+n;++i)//擴充套件in 
    		arr1[i]=arr1[i-n];
    	for(int i=1;i<=n;++i)//輸入out 
    		cin>>arr2[i];
    	for(int i=n+1;i<=n+n;++i)//擴充套件out 
    		arr2[i]=arr2[i-n];
    	
    	while(que.size())
    		que.pop_back();
    	bool flag=0;
    	for(int i=1;i<=n+n;++i)
    	{
    		if(sum+arr1[i]-arr2[i]>=0)//符合要求 
    		{
    			que.push_back(i);
    			sum=sum+arr1[i]-arr2[i];
    			if(que.size()>=n)
    			{
    				flag=1;
    				cout<<que.front()<<endl;
    				break;
				}
			}
			else//不符合要求 
			{
				while(sum+arr1[i]-arr2[i]<0&&que.size()>0)
				{
					sum=sum-arr1[que.front()]+arr2[que.front()];
					que.pop_front();
				}
				//直到符合要求||從新開始 
				if(sum+arr1[i]-arr2[i]>=0)
				{
					que.push_back(i);
					sum=sum+arr1[i]-arr2[i];
				}
				if(que.size()>=n)
				{
					flag=1;
					cout<<que.front()<<endl;
					break;
				}
			}
		}
		if(!flag)
			cout<<-1<<endl;
	}
	return 0;
}