UVa11093

 //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>

 using namespace std;

 const int maxn = ;

 int t;

 int n;

 int icase;

 int a[maxn+];

 int b[maxn+];

 int main(){
         scanf("%d",&t);
         while(t--){
             scanf("%d",&n);
             memset(a, , sizeof(a));
             memset(b, , sizeof(b));
             for(int i = ; i <= n; ++i){
                 scanf("%d",&a[i]);
             }
             for(int i = ; i <= n; ++i){
                 scanf("%d",&b[i]);
             }
             if(n == ){
                 if(a[] >= b[]){
                     printf("Case %d: Possible from station %d\n",++icase,);
                 }else {
                     printf("Case %d: Not possible\n",++icase);
                 }
                 continue;
             }
             int cur = ;
             int i = ;
             int cnt = ;
             int fuel = ;
             int flag = ;
             while(cnt != n){
                 fuel += a[i];
                 if(fuel >= b[i]){
                     fuel -= b[i];
                     ++i;
                     ++cnt;
                     if(i > n){
                         i = ;
                         flag = ;
                     }
                 } else {
                     if(flag == ){
                         flag = ;
                         break;
                     }
                     ++i;
                     if(i > n){
                         i = ;
                         flag = ;
                     }
                     cur = i;
                     cnt = ;
                     fuel = ;
                 }
             }
             if(flag == ){
                 printf("Case %d: Not possible\n",++icase);
             } else {
                 printf("Case %d: Possible from station %d\n",++icase,cur);
             }
         }
 }