HDU - 3450 Counting Sequences

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题意：求个数大于等于2的序列，要求每相邻的两个的大于d,求满足的个数

思路：同样是树状数组的应用，跟前面两题类似，求每次加入的a[i],先求范围在[a[i]-d,a[i]+d]的个数，再加到a[i]上，加一加的是本身，还有要注意的是，要减去1个的情况，跟前面两题不一样，

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 100005;
const int MOD = 9901;int n,arr[MAXN],k;
int brr[MAXN],cnt,f[MAXN];
int crr[MAXN];int lowbit(int x){return x&(-x);
}void update(int x,int d){while (x < MAXN){crr[x]  = (crr[x]+d) % MOD;x += lowbit(x);}
}int sum(int x){int ans = 0;while (x > 0){ans = (ans + crr[x])%MOD;x -= lowbit(x);}return ans;
}int main(){while (scanf("%d%d",&n,&k) != EOF){cnt = 1;memset(crr,0,sizeof(crr));for (int i = 0; i < n; i++){scanf("%d",&arr[i]);brr[i] = arr[i];}sort(brr,brr+n);f[0] = brr[0];for (int i = 1; i < n; i++)if (brr[i] != brr[i-1])f[cnt++] = brr[i];f[cnt++] = INF;	for (int i = 0; i < n; i++){int x = lower_bound(f,f+cnt,arr[i]) - f + 1;int y = upper_bound(f,f+cnt,arr[i]+k) - f;int z = lower_bound(f,f+cnt,arr[i]-k) - f + 1;int val = sum(y) - sum(z-1) + 1;val = (val%MOD+MOD) % MOD;update(x,val);}int ans = sum(cnt);ans = ((ans-n)%MOD+MOD)%MOD;cout << ans << endl;}return 0;
}

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