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 = 100001;

int t;

int n;

int icase;

int a[maxn+10];

int b[maxn+10];

int main(){
        scanf("%d",&t);
        while(t--){
            scanf("%d",&n);
            memset(a, 0, sizeof(a));
            memset(b, 0, sizeof(b));
            for(int i = 1; i <= n; ++i){
                scanf("%d",&a[i]);
            }
            for(int i = 1; i <= n; ++i){
                scanf("%d",&b[i]);
            }
            if(n == 1){
                if(a[1] >= b[1]){
                    printf("Case %d: Possible from station %d
",++icase,1);
                }else {
                    printf("Case %d: Not possible
",++icase);
                }
                continue;
            }
            int cur = 1;
            int i = 1;
            int cnt = 0;
            int fuel = 0;
            int flag = 0;
            while(cnt != n){
                fuel += a[i];
                if(fuel >= b[i]){
                    fuel -= b[i];
                    ++i;
                    ++cnt;
                    if(i > n){
                        i = 1;
                        flag = 1;
                    }
                } else {
                    if(flag == 1){
                        flag = 2;
                        break;
                    }
                    ++i;
                    if(i > n){
                        i = 1;
                        flag = 1;
                    }
                    cur = i;
                    cnt = 0;
                    fuel = 0;
                }
            }
            if(flag == 2){
                printf("Case %d: Not possible
",++icase);
            } else {
                printf("Case %d: Possible from station %d
",++icase,cur);
            }
        }
}