hdu3450——Counting Sequences

Counting Sequences

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 1872    Accepted Submission(s): 635

Problem Description

For a set of sequences of integers｛a1，a2，a3，...an｝, we define a sequence｛ai1,ai2,ai3...aik｝in which 1<=i1<i2<i3<...<ik<=n, as the sub-sequence of ｛a1，a2，a3，...an｝. It is quite obvious that a sequence with the length n has 2^n sub-sequences. And for a sub-sequence｛ai1,ai2,ai3...aik｝，if it matches the following qualities: k >= 2, and the neighboring 2 elements have the difference not larger than d, it will be defined as a Perfect Sub-sequence. Now given an integer sequence, calculate the number of its perfect sub-sequence.

 

Input

Multiple test cases The first line will contain 2 integers n, d（2<=n<=100000，1<=d=<=10000000） The second line n integers, representing the suquence

 

Output

The number of Perfect Sub-sequences mod 9901

 

Sample Input

4 21 3 7 5

 

Sample Output

4

 

Source

2010 ACM-ICPC Multi-University Training Contest（2）——Host by BUPT

 

Recommend

线段树+dp，dp方程很简单,dp[i]表示以第i个元素结尾的完美序列的个数
dp[i] += dp[j] + 1(j < i)

先把数据离散化，然后按顺序插入到线段树上，维护区间和

#include <map>#include <set>#include <list>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 100010;__int64 dp[N];int xis[N];int arr[N];int cnt;struct node{int l, r;__int64 sum;}tree[N << 2];int BinSearch(int val){int l = 1, r = cnt, mid, ans;while (l <= r){mid = (l + r) >> 1;if (xis[mid] == val){ans = mid;break;}else if (xis[mid] > val){r = mid - 1;}else{l = mid + 1;}}return ans;}int BinSearch_left(int val){int l = 1;int r = cnt, mid, ans;while (l <= r){mid = (l + r) >> 1;if (xis[mid] >= val){ans = mid;r = mid - 1;}else{l = mid + 1;}}return ans;}int BinSearch_right(int val){int l = 1;int r = cnt, mid, ans;while (l <= r){mid = (l + r) >> 1;if (xis[mid] <= val){ans = mid;l = mid + 1;}else{r = mid - 1;}}return ans;}void build(int p, int l, int r){tree[p].l = l;tree[p].r = r;tree[p].sum = 0;if (l == r){return;}int mid = (l + r) >> 1;build(p << 1, l, mid);build(p << 1 | 1, mid + 1, r);}void update(int p, int pos, __int64 val){if (tree[p].l == tree[p].r){tree[p].sum += val;return;}int mid = (tree[p].l + tree[p].r) >> 1;if (pos <= mid){update(p << 1, pos, val);}else{update(p << 1 | 1, pos, val);}tree[p].sum = tree[p << 1].sum + tree[p << 1 | 1].sum;tree[p].sum %= 9901;}__int64 query(int p, int l, int r){if (tree[p].l >= l && tree[p].r <= r){return tree[p].sum;}int mid = (tree[p].l + tree[p].r) >> 1;if (r <= mid){return query(p << 1, l, r);}else if (l > mid){return query(p << 1 | 1, l, r);}else{return query(p << 1, l, mid) + query(p << 1 | 1, mid + 1, r);}}int main(){int n, d;while(~scanf("%d%d", &n, &d)){memset (dp, 0, sizeof(dp));__int64 ans = 0;for (int i = 1; i <= n; ++i){scanf("%d", &arr[i]);xis[i] = arr[i];}sort(xis + 1, xis + n + 1);cnt = unique(xis + 1, xis + n + 1) - xis - 1;build(1, 1, cnt);dp[1] = 0;update(1, BinSearch(arr[1]), dp[1] + 1);int l, r;for (int i = 2; i <= n; i++){l = BinSearch_left(arr[i] - d);r = BinSearch_right(arr[i] + d);dp[i] = query(1, l, r);dp[i] %= 9901;int x = BinSearch(arr[i]);update(1, x, dp[i] + 1);ans += dp[i];ans %= 9901;}printf("%I64d\n", ans);}return 0;}