2016 UESTC Training for Dynamic Programming N - 柱爷与子序列 这题和N题有些相似之处、用了树状数组

N - 柱爷与子序列

所谓完美子序列，还需满足以下条件：

m≥2

|Abi−Abi+1|≤k,1≤i<m

Input

Output

Sample input and output

4 2
1 3 7 5

4

Hint

Source

My Solution

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 100000 + 8, HASH = 1000000009;
int a[maxn], b[maxn], nn;
//LL dp[maxn];

LL Tree[maxn];

inline int lowbit(int x)
{
return (x&-x);
}

void add(int x, LL value)
{
for(int i = x; i <= nn; i += lowbit(i))
Tree[i] = (Tree[i] + value) % HASH;
}

LL get(int x)
{
LL sum = 0;
for(int i = x; i; i -= lowbit(i))
sum =(sum + Tree[i]) % HASH;
return sum;
}
//!WA7 然后直接给中间过程已经树状数组里面也直接全部取模了☺☺
int main()
{
#ifdef LOCAL
freopen("a.txt", "r", stdin);
#endif // LOCAL
int n, k, x, l, r;
long long dp;  //dp[maxn] ==>> dp
scanf("%d%d", &n, &k);
for(int i = 0; i < n; i++)
scanf("%d", &a[i]);
memcpy(b, a, sizeof a);
sort(a, a + n);
nn = unique(a, a + n) - a;

memset(Tree, 0, sizeof Tree);
for(int i = 0; i < n; i++){                            //!这里坐标统一为 1 ~ nn 了
x = lower_bound(a, a + nn, b[i]) - a + 1;
l = lower_bound(a, a + nn, b[i] - k) - a + 1;
r = upper_bound(a, a + nn, b[i] + k) - a;          //返回最后一个b[i]的后面一个位置。
//cout<<"x = "<<x<<" l = "<<l<<" r = "<<r<<endl;
dp = (get(r) - get(l-1))%HASH;                            //有一个 l - 1的所以还是 1 ~ nn 好
add(x, dp+1);                                      //修改的时候不能是0的不然 0&-0一直这样会死循环，当然，这里x 不会是0
}
printf("%lld", (get(nn) - n +HASH)%HASH); //ans 不是 dp[n-1] ☺☺
return 0;
}