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();
}