codeforces Round #259(div2) B解题报告_html/css_WEB-ITnose

B. Little Pony and Sort by Shift

time limit per test

memory limit per test

input

output

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?

Input

The first line contains an integer n (2?≤?n?≤?105). The second line contains n integer numbers a1,?a2,?...,?an (1?≤?ai?≤?105).

Output

If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.

立即学习“前端免费学习笔记（深入）”；

Sample test(s)

input

22 1

output

input

31 3 2

output

-1

input

21 2

output

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"。

猫眼课题宝

5分钟定创新选题，3步生成高质量标书！

85

查看详情

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]

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;	else		for (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);	else		printf("-1\n");}int main() {	init();	solve();}