CodeForces - 831A Unimodal Array 模拟
2018-01-24 23:29
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A. Unimodal Array
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Array of integers is unimodal, if:
- it is strictly increasing in the beginning;
- after that it is constant;
- after that it is strictly decreasing.
The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.
For example, the following three arrays are unimodal: [5, 7, 11, 11, 2, 1], [4, 4, 2], [7], but the following three are not unimodal: [5, 5, 6, 6, 1], [1, 2, 1, 2], [4, 5, 5, 6].
Write a program that checks if an array is unimodal.
InputThe first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the array.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1 000) — the elements of the array.
OutputPrint "YES" if the given array is unimodal. Otherwise, print "NO".
You can output each letter in any case (upper or lower).
Examples Input6Output
1 5 5 5 4 2
YESInput
5Output
10 20 30 20 10
YESInput
4Output
1 2 1 2
NOInput
7Output
3 3 3 3 3 3 3
YES
直接按题意进行判断
#include<stdio.h> bool flag; bool flag1; bool flag2; int main() { int n; int a[105]; while (~scanf("%d", &n)) { flag = true; flag1 = true; flag2 = true; for (int i = 0; i < n; i++) { scanf("%d", &a[i]); } if (n == 1) { printf("YES\n"); continue; } for (int i = 1; i < n; i++) { if (a[i] == a[i - 1]&&flag1 ) { flag = false; } else if (a[i]<a[i - 1]) { flag = false; flag1 = false; } else if (a[i] > a[i - 1] && (!flag||!flag1)) { flag2 = false; break; } else if (a[i] == a[i - 1] &&!flag1) { flag2 = false; } } if (!flag2) { printf("NO\n"); } else { printf("YES\n"); } } return 0; }
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