codeforces Round #259(div2) B解题报告
来源:互联网 发布:白夜追凶网络剧 编辑:程序博客网 时间:2024/06/05 03:36
One day, Twilight Sparkle is interested in how to sort a sequence of integers a1, a2, ..., an in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:
Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?
The first line contains an integer n (2 ≤ n ≤ 105). The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105).
If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.
22 1
1
31 3 2
-1
21 2
0
题目大意:
给出N个数字,可以每一次将最后一个数字移动到最前面,要求最终状态是一个单调非递减的序列,求最少需要花多少次操作。如若无法达到目标则输出“-1"。
解法:
也是一道很easy的编程基础题,找出两队单调非递减序列,分别为1~x 和 x+1~y,判断这两队是否覆盖整串数字,且a[n] <= a[1]。
更简单的一种做法就是,将a[1]~a[n]复制一遍,拓展到a[1]~a[2*n],然后在1 ~ 2*n里面找,是否有一串单调不递减的个数为n的序列。
代码:
#include <cstdio>#define N_max 123456int n, x, y, cnt;int a[N_max];void init() {scanf("%d", &n);for (int i = 1; i <= n; i++) scanf("%d", &a[i]);}void solve() {for (int i = 1; i <= n; i++)if (a[i] > a[i+1]) {x = i;break;}if (x == n)y = n;elsefor (int i = x+1; i <= n; i++)if (a[i] > a[i+1]) {y = i;break;}if (x == n)printf("0\n");else if (y == n && a[y] <= a[1])printf("%d\n", y-x);elseprintf("-1\n");}int main() {init();solve();}
- codeforces Round #259(div2) B解题报告
- codeforces Round #237(div2) B解题报告
- codeforces Round #238(div2) B解题报告
- codeforces Round #241(div2) B解题报告
- codeforces Round #269(div2) B解题报告
- codeforces Round #267(div2) B解题报告
- codeforces Round #271(div2) B解题报告
- codeforces Round #272(div2) B解题报告
- codeforces Round #264(div2) B解题报告
- codeforces Round #263(div2) B解题报告
- codeforces Round #273(div2) B解题报告
- codeforces Round #274(div2) B解题报告
- codeforces Round #261(div2) B解题报告
- codeforces Round #260(div2) B解题报告
- codeforces Round #275(div2) B解题报告
- codeforces Round #258(div2) B解题报告
- codeforces Round #259(div2) A解题报告
- codeforces Round #259(div2) C解题报告
- 字符串的相似度计算
- phpcmsv9修改fckeditor编辑器为ueditor编辑器
- js如何判断一个对象{}是否为空对象,没有任何属性
- C语言第八天
- [django]django-orm中F对象的使用
- codeforces Round #259(div2) B解题报告
- 背包问题
- python中子类实例化调用父类方法
- 阿里云部署Docker(9)----Dockerfile脚本定制你的镜像
- org.hibernate.PropertyAccessException: Null value was assigned to a property of primitive type sette
- 常用 Block 定义
- 梦幻
- Git常用命令以及用法
- OC_关于NSString的使用