CF279C Ladder 简单DP
2016-03-21 13:06
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L[x]代表不递增序列,从x位置向左最远能延伸到的位置
R[x]代表不递减序列,从x位置向右最远能延伸到的位置
给出x,y
只用判断R[x]是否>=L[y]即可。
一看就感觉是水题。
C. Ladder
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You've got an array, consisting of n integers a1, a2, ..., an.
Also, you've got m queries, the i-th
query is described by two integers li, ri.
Numbersli, ri define
a subsegment of the original array, that is, the sequence of numbers ali, ali + 1, ali + 2, ..., ari.
For each query you should check whether the corresponding segment is a ladder.
A ladder is a sequence of integers b1, b2, ..., bk,
such that it first doesn't decrease, then doesn't increase. In other words, there is such integer x(1 ≤ x ≤ k),
that the following inequation fulfills: b1 ≤ b2 ≤ ... ≤ bx ≥ bx + 1 ≥ bx + 2... ≥ bk.
Note that the non-decreasing and the non-increasing sequences are also considered ladders.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) —
the number of array elements and the number of queries. The second line contains the sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 109),
where number ai stands
for the i-th array element.
The following m lines contain the description of the queries. The i-th
line contains the description of the i-th query, consisting of two integers li, ri(1 ≤ li ≤ ri ≤ n) —
the boundaries of the subsegment of the initial array.
The numbers in the lines are separated by single spaces.
Output
Print m lines, in the i-th
line print word "Yes" (without the quotes), if the subsegment that corresponds to the i-th
query is the ladder, or word "No" (without the quotes) otherwise.
Examples
input
output
R[x]代表不递减序列,从x位置向右最远能延伸到的位置
给出x,y
只用判断R[x]是否>=L[y]即可。
一看就感觉是水题。
C. Ladder
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You've got an array, consisting of n integers a1, a2, ..., an.
Also, you've got m queries, the i-th
query is described by two integers li, ri.
Numbersli, ri define
a subsegment of the original array, that is, the sequence of numbers ali, ali + 1, ali + 2, ..., ari.
For each query you should check whether the corresponding segment is a ladder.
A ladder is a sequence of integers b1, b2, ..., bk,
such that it first doesn't decrease, then doesn't increase. In other words, there is such integer x(1 ≤ x ≤ k),
that the following inequation fulfills: b1 ≤ b2 ≤ ... ≤ bx ≥ bx + 1 ≥ bx + 2... ≥ bk.
Note that the non-decreasing and the non-increasing sequences are also considered ladders.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) —
the number of array elements and the number of queries. The second line contains the sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 109),
where number ai stands
for the i-th array element.
The following m lines contain the description of the queries. The i-th
line contains the description of the i-th query, consisting of two integers li, ri(1 ≤ li ≤ ri ≤ n) —
the boundaries of the subsegment of the initial array.
The numbers in the lines are separated by single spaces.
Output
Print m lines, in the i-th
line print word "Yes" (without the quotes), if the subsegment that corresponds to the i-th
query is the ladder, or word "No" (without the quotes) otherwise.
Examples
input
8 6 1 2 1 3 3 5 2 1 1 3 2 3 2 4 8 8 1 4 5 8
output
Yes Yes No Yes No Yes
#include<cstdio> #include<string> #include<cstring> #include<iostream> #include<cmath> #include<algorithm> using namespace std; typedef long long ll; const int INF =0x3f3f3f3f; const int maxn= 100000 ; int a[maxn+5]; int R[maxn+5]; int L[maxn+5]; int n,m; int main() { while(~scanf("%d%d",&n,&m)) { for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } R =n; for(int i=n-1;i>=1;i--) { if(a[i+1]<a[i]) R[i]=i; else { R[i]=R[i+1]; } } L[1]=1; for(int i=2;i<=n;i++) { if(a[i-1]<a[i]) L[i]=i; else { L[i]=L[i-1]; } } int le,ri; for(int i=1;i<=m;i++) { scanf("%d%d",&le,&ri); puts(R[le]>=L[ri]?"Yes":"No"); } } return 0; }
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