HDU - 3030 Increasing Speed Limits

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题意：求有长为m的序列生成的长度为n的序列的上升子序列的个数

思路：生成完长为n的序列后，首先我们想到（nlogn）的求上升序列的方法，然后再这个基础上改进，每插入一个的时候，我们可以得到左边的小于它的个数，然后我们就可以得新增加的上升子序列是sum(id-1)+1,然后更新树状数组

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long 
using namespace std;
const int MOD = 1000000007;
const int MAXN = 500010;struct Node{int id;ll x;
}node[MAXN];
ll c[MAXN],a[MAXN],val[MAXN];
ll x,y,z;
int b[MAXN],n,m;bool cmp(Node a,Node b){return a.x < b.x;
}int lowbit(int x){return x&(-x);
}void update(int x,ll d){while (x <= n){c[x] += d;x += lowbit(x);c[x] %= MOD;}
}ll sum(int x){ll ans = 0;while (x > 0){ans += c[x];ans %= MOD;x -= lowbit(x);}return ans;
}int main(){int t,cas=1;scanf("%d",&t);while (t--){scanf("%d%d%lld%lld%lld",&n,&m,&x,&y,&z);for (int i = 0; i < m; i++)scanf("%lld",&a[i]);for (int i = 0; i < n; i++){b[i+1] = val[i+1] = a[i%m];a[i%m] = (x*a[i%m]+y*(i+1))%z;}sort(b+1,b+1+n);memset(c,0,sizeof(c));ll ans = 0;node[0].id = -1;for (int i = 1; i <= n; i++){int id = lower_bound(b+1,b+1+n,val[i])-b;ll tot = sum(id-1);ans += tot+1;ans %= MOD;update(id,tot+1);}printf("Case #%d: ",cas++);cout << ans << endl;}return 0;
}

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