题干：
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.
Input
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
Output
For each test case, output the start city you should choose.
Sample Input
2 3 0 3 4 5 5 4 3 3 100 -3 -4 -5 30 40 50
Sample Output
2 -1
Hint
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.
题目大意：
有n个城市围成一圈 编号为1-n。第一次到达第i个城市需要获得ai金币，同时如果想从当前城市前往其他城市，需要花费bi金币。起初有c金币。可以任意选择一个城市作为起点，问能否从一个城市出发环游一圈？【起点等于经过了两次】
解题报告：
刚开始读错题了，以为是从每个点可以走到任意一个合法的点，但是其实是1~n围成一个圈，所以必须顺着走。这就更简单了。
数据水了，，按题目意思来说，应该特判n==1这种情况？而且尺取的话少写一句维护左端点也能过？
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e6 + 5;
int n;
ll c,a[MAX],b[MAX];
int main()
{int t;cin>>t;while(t--) {scanf("%d%lld",&n,&c);for(int i = 1; i<=n; i++) cin>>a[i],a[i+n] = a[i];for(int i = 1; i<=n; i++) cin>>b[i],b[i+n] = b[i];ll x = 0;int l = 1,ans = -1;for(int r = 1; r<=2*n; r++) {x += a[r]-b[r];while(x < -c && l <= r) {x -= a[l]-b[l];l++; }while(r-l+1 > n) x -= a[l]-b[l],l--; //为什么少写这一句也能过？if(r-l+1 == n && x >= -c) {ans = l;break;}}printf("%d\n",ans);}return 0 ;
}线段树代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e6 + 5;
struct TREE {int l,r;ll minn;
} tr[MAX<<2];
int n;
ll c;
ll a[MAX],b[MAX];
void pushup(int cur) {tr[cur].minn = min(tr[cur*2].minn,tr[cur*2+1].minn);
}
void build(int l,int r,int cur) {tr[cur].l = l;tr[cur].r = r;if(l == r) {tr[cur].minn = a[l];return;}int m = (l+r)>>1;build(l,m,cur*2);build(m+1,r,cur*2+1);pushup(cur);
}
ll query(int pl,int pr,int cur) {if(pl <= tr[cur].l && pr >= tr[cur].r) {return tr[cur].minn;} ll res = 1e15;if(pl <= tr[cur*2].r) res = query(pl,pr,cur*2);if(pr >= tr[cur*2+1].l) res = min(res,query(pl,pr,cur*2+1));return res;
}int main()
{int T;cin>>T;while(T--) {scanf("%d%lld",&n,&c);for(int i = 1; i<=n; i++) scanf("%lld",a+i),a[i+n] = a[i];for(int i = 1; i<=n; i++) scanf("%lld",b+i),b[i+n] = b[i];if(n == 1) {if(c-a[1]>=0) printf("1\n");else printf("-1\n");continue; }for(int i = 1; i<=2*n; i++) a[i] -= b[i],a[i]+=a[i-1];build(1,2*n,1);int ans = -1;for(int i = 1; i<=n; i++) {ll x = query(i,i+n-1,1)-a[i-1];if(x >= -c) {ans = i;break;}}printf("%d\n",ans);} return 0 ;
}
