UVa 11093

循环遍历数组即可，注意临界条件的判断。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 102400;

int p[maxn], q[maxn];

int main()
{
    //freopen("in.txt", "r", stdin);
    int T; cin >> T;
    for(int t = 1; t <= T; ++t){
        int n; cin >> n;
        for(int i = 0; i < n; ++i)
            cin >> p[i];
        for(int i = 0; i < n; ++i)
            cin >> q[i];
        bool ok = true;
        int beg = 0;
        for(int i = 0, j = 0; i < n; ++i){
            int sum = 0;
            for(j = 0; j < n; ++j){
                int cur = (i+j)%n;
                sum += p[cur];
                if(sum >= q[cur])
                    sum -= q[cur];
                else{
                    if(cur < i || cur == n-1) {ok = false; break;}
                    else {i = cur; beg = cur + 1; break;}
                }
            }
            if(j == n || !ok) break;
        }
        if(!ok) printf("Case %d: Not possible
", t);
        else printf("Case %d: Possible from station %d
", t, beg + 1);
    }
    return 0;
}