摘要: int n, a[5000]; int main(){ scanf("%d", &n); for (int i = 1; i > a[i]; ll cur = 0, ans = 0, l, r, tl = 1; for (int i = 1; i ans){ ans = cur; r = i; ... 阅读全文
posted @ 2018-06-02 15:48 心照不必宣 阅读(653) 评论(0) 推荐(0)