Codeforces Round #258 (Div. 2) B. Sort the Array
来源:互联网 发布:青年文明号在线网络办公协同系统 编辑:程序博客网 时间:2024/05/10 03:48
Being a programmer, you like arrays a lot. For your birthday, your friends have given you an arraya consisting of ndistinct integers.
Unfortunately, the size of a is too small. You want a bigger array! Your friends agree to give you a bigger array, but only if you are able to answer the following question correctly: is it possible to sort the arraya (in increasing order) by reversing exactly one segment of a? See definitions of segment and reversing in the notes.
The first line of the input contains an integer n (1 ≤ n ≤ 105) — the size of arraya.
The second line contains n distinct space-separated integers:a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109).
Print "yes" or "no" (without quotes), depending on the answer.
If your answer is "yes", then also print two space-separated integers denoting start and end (start must not be greater than end) indices of the segment to be reversed. If there are multiple ways of selecting these indices, print any of them.
33 2 1
yes1 3
42 1 3 4
yes1 2
43 1 2 4
no
21 2
yes1 1
题目大意:
一个序列求是否可以反转某一部分,使得的序列为递增的,注意一句话then also print two space-separated integers denoting start and end,输出的是位置,而不是当前
的数值。并若是反转有reverse()函数,用sort()WA。。
思路:从左到右找到比前一个数小的数的下标,然后从右向左找比后一个数大的下标。找到便break
#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<cmath>#include<algorithm>#define LL int#define inf 0x3f3f3f3fusing namespace std;LL a[1000001],b[1000001];int main(){ LL n,m,i,j,k,x,y; while(scanf("%d",&n)!=EOF) { x=y=1; for(i=1; i<=n; i++) scanf("%d",&a[i]); for(i=1; i<n; i++) { if(a[i]>a[i+1]) { x=i; break; } } for(i=n; i>=2; i--) if(a[i]<a[i-1]) { y=i; break; } reverse(a+x,a+y+1); bool vis=false; for(i=1; i<=n-1; i++) { if(a[i]>a[i+1]) { vis=true; } } if(vis) printf("no\n"); else printf("yes\n%d %d\n",x,y); } return 0;}
- Codeforces Round #258 (Div. 2) 2B Sort the Array
- Codeforces Round #258 (Div. 2) B. Sort the Array
- Codeforces Round #258 (Div. 2) B. Sort the Array
- Codeforces Round #258 (Div. 2) B. Sort the Array
- Codeforces Round #258 (Div. 2) B. Sort the Array
- Codeforces Round #258 (Div. 2) B. Sort the Array
- Codeforces Round #258 (Div. 2)——B. Sort the Array
- Codeforces Round #258 (Div. 2) B. Sort the Array (模拟)
- Codeforces Round #258 (Div. 2) B. Sort the Array(简单题)
- Codeforces Round #258 (Div. 2)——B. Sort the Array(STL-reverse)
- Codeforces #258 (Div. 2) B. Sort the Array
- Codeforces Div. 2 #258-B. Sort the Array
- Codeforces Round #258 (Div. 2/B)/Codeforces451B_Sort the Array
- Codeforces Round #138 (Div. 2) B. Array
- Codeforces Round #312 (Div. 2) B. Amr and The Large Array
- Codeforces Round #312 (Div. 2) B. Amr and The Large Array
- Codeforces Round #312 (Div. 2) B. Amr and The Large Array
- Codeforces Round #312 (Div. 2)-B. Amr and The Large Array-暴力
- Kafka 配置参数(非常好的总结)
- 八皇后问题 (白书P192)
- 2012 Multi-University Training Contest 5 Problem F Permutation(HDU4345)
- 控制CPU使用率
- 黑马程序员——集合框架的使用及其注意事项
- Codeforces Round #258 (Div. 2) B. Sort the Array
- how-to: resolve crontab does not work
- 遗传算法入门
- 网络解析数据
- hdu1997汉诺塔VII
- Single boy 你距离女神,只差一个小恩爱
- WIN7+VS2012+COCOS2D-X 3.7创建的项目移植到iphone IOS步骤
- Event C++初识
- FFmpeg深入分析之零-基础