首页
移动开发
物联网
服务端
编程语言
企业开发
数据库
业界资讯
其他
搜索
Codeforces Round #618 (Div. 2) A. Non-zero
其他
2020-02-10 15:00:49
阅读次数: 0
题意:使数组和不为0并且积不为0,可以使ai=ai+1,求最少操作数
题解:记录0的个数以及数组和
如果0的个数+数组和==0 输出0的个数+1
否则输出0的个数
猜你喜欢
转载自
www.cnblogs.com/RE-TLE/p/12290885.html
Codeforces Round #618 (Div. 2)A. Non-zero
Codeforces Round #618 (Div. 2) A. Non-zero
【Codeforces Round#618 (Div. 2)】A. Non-zero 题解
Codeforces Round #618 (Div. 2)A、Non-zero
Codeforces Round #618 (Div. 2)-Non-zero
Codeforces Round #618 (Div. 2) 题解
Codeforces Round #618 (Div. 2)
Codeforces Round #618 (Div. 2)题解
Codeforces Round #618 (Div. 2) 【A - E】
Codeforces Round #618 (Div. 2) E
Codeforces Round #618 (Div. 2)(A~E)
Codeforces Round #618 (Div. 2)(题解)
Codeforces Round #618 (Div. 2)补题
Codeforces Round #618 (Div. 2)(A~E)
Codeforces Round #572 (Div. 2) A.
Codeforces Round #618 (Div. 2) (位运算计数)
Codeforces Round #618 (Div. 2) E. Water Balance
Codeforces Round #618 (Div. 2) D. Aerodynamic
Codeforces Round #618 (Div. 2) B. Assigning to Classes
Codeforces Round #618 (Div. 2)-B. Assigning to Classes
Codeforces Round #618 (Div. 2)B、Assigning to Classes
Codeforces Round #618 (Div. 2)C、Anu Has a Function
Codeforces Round #618 (Div. 2) 小号上紫之路
Codeforces Round #618 (Div. 2) C题
Codeforces Round #618 (Div. 2)解题报告
Codeforces Round #618 (Div. 2)全解
Codeforces Round #618 (Div. 2)A-E题解
Codeforces Round #618 (Div. 2) 题解报告
A. Road To Zero(贪心) Educational Codeforces Round 86 (Rated for Div. 2)
A. Road To Zero(水 Educational Codeforces Round 86 Rated for Div. 2)
今日推荐
周排行
LRU cache算法
windows10, 自带的OpenSSH, key权限问题, 文件权限问题
测试用例书写方法
HIVE-默认分隔符的(linux系统的特殊字符)查看,输入和修改
最贵的AMD 7nm显卡来了!这设计 够狂野
java多线程简单demo
[ 转载 ]在Android系统上使用busybox——最简单的方法
QT connect学习
BFSIFT算法分析
Xcode10:library not found for -lstdc++.6.0.9 临时解决
每日归档
更多
2024-08-06(0)
2024-08-05(0)
2024-08-04(0)
2024-08-03(0)
2024-08-02(0)
2024-08-01(0)
2024-07-31(0)
2024-07-30(0)
2024-07-29(0)
2024-07-28(0)