HDU - 3030 Increasing Speed Limits

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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HDU - 3030 Increasing Speed Limits

原文：http://blog.csdn.net/u011345136/article/details/23535967