//当汽车从第i个加油站到第j个加油站无法继续走下去的时候，这时候[i,j]区间的所有加油站都无法作为起点，因为当我们到第k个加油站的时候，起码是带着>=0的油去的，现在不带油直接从第k个开始肯定更不行了。
 1 #include<bits/stdc++.h> 2  3 using namespace std; 4  5 const int maxn = 100001; 6  7 int t; 8  9 int n;10 11 int icase;12 13 int a[maxn+10];14 15 int b[maxn+10];16 17 int main(){18         scanf("%d",&t);19         while(t--){20             scanf("%d",&n);21             memset(a, 0, sizeof(a));22             memset(b, 0, sizeof(b));23             for(int i = 1; i <= n; ++i){24                 scanf("%d",&a[i]);25             }26             for(int i = 1; i <= n; ++i){27                 scanf("%d",&b[i]);28             }29             if(n == 1){30                 if(a[1] >= b[1]){31                     printf("Case %d: Possible from station %d\n",++icase,1);32                 }else {33                     printf("Case %d: Not possible\n",++icase);34                 }35                 continue;36             }37             int cur = 1;38             int i = 1;39             int cnt = 0;40             int fuel = 0;41             int flag = 0;42             while(cnt != n){43                 fuel += a[i];44                 if(fuel >= b[i]){45                     fuel -= b[i];46                     ++i;47                     ++cnt;48                     if(i > n){49                         i = 1;50                         flag = 1;51                     }52                 } else {53                     if(flag == 1){54                         flag = 2;55                         break;56                     }57                     ++i;58                     if(i > n){59                         i = 1;60                         flag = 1;61                     }62                     cur = i;63                     cnt = 0;64                     fuel = 0;65                 }66             }67             if(flag == 2){68                 printf("Case %d: Not possible\n",++icase);69             } else {70                 printf("Case %d: Possible from station %d\n",++icase,cur);71             }72         }73 }