hdu 3030 Increasing Speed Limits

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3030

题目大意: 通过题目描述的方法产生一个正整数列,求递增非连续子数列的个数.

思路:dp[x]=dp[i]+...+dp[j]+1 (B[]为产生的正整数列,B[i]~B[j]的值小于B[x] && i,j<x,1为B[x]本身).求和可以用线段树和树状数组.

代码:

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

#define ull unsigned __int64
#define ll __int64
//#define ull unsigned long long
//#define ll long long
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define middle (l+r)>>1
#define MOD 1000000007
#define esp (1e-10)
const int INF=0x3F3F3F3F;
//const double pi=acos(-1.0);
const int N=500010;
int n,m;
ll a[N],b[N],sum[N],hash[N];

void mod(ll& x){x= x<MOD? x:x%MOD;}

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

void Add(int pos,ll c){
    while(pos<=m) mod(sum[pos]+=c),pos+=lowbit(pos);
}

ll Sum(int pos){
    ll r=0;
    while(pos>0) mod(r+=sum[pos]),pos-=lowbit(pos);
    return r;
}

int bs(ll key,int size,ll A[]){
    int l=0,r=size-1,mid;
    while(l<=r){
        mid=middle;
        if(key>A[mid]) l=mid+1;
        else if(key<A[mid]) r=mid-1;
        else return mid;
    }
    return -1;
}

int main(){
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    int i,j,k,pos;
    ll ret,X,Y,Z,tmp;
    char op[2];
    int T,cas;scanf("%d",&T);for(cas=1;cas<=T;cas++){
        scanf("%d%d%I64d%I64d%I64d",&n,&m,&X,&Y,&Z);
        for(i=0;i<m;i++) scanf("%I64d",&a[i]);
        for(i=0;i<n;i++){
            j=i%m;hash[i]=b[i]=a[j];
            a[j]=(X*a[j]+Y*(i+1))%Z;
        }
        sort(hash,hash+n);
        for(i=m=1;i<n;i++) if(hash[i]!=hash[i-1]) hash[m++]=hash[i];
        memset(sum,0,sizeof(sum));
        for(i=0,ret=0;i<n;i++){
            pos=bs(b[i],m,hash)+1;
            tmp=Sum(pos)+1;
            mod(ret+=tmp);
            Add(pos+1,tmp);
        }
        printf("Case #%d: %I64d\n",cas,ret);
    }
    return 0;
}