codeforcesRound#259(div2)B解题报告

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 a 1 ,? a 2 ,?...,?

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.

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();
}</cstdio>