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codeforces Round #259(div2) B解题报告

2024-07-29 03:29:58   224人阅读

B. Little Pony and Sort by Shift
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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:

a1,?a2,?...,?an?→?an,?a1,?a2,?...,?an?-?1.

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.

Sample test(s)
input
2
2 1
output
1
input
3
1 3 2
output
-1
input
2
1 2
output
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 123456

int 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();
}

codeforces Round #259(div2) B解题报告


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