hdu 3030 Increasing Speed Limits

这道题和上篇介绍的hdu 3450 是同一个类型的题。

有点不相同的地方是这道题可以包含长度为1的序列！

方法：树状数组+DP

code:

View Code

 1 # include<stdio.h>
 2 # include<string.h>
 3 # include<stdlib.h>
 4 # define N 500005
 5 # define mod 1000000007
 6 __int64 a[N],b[N],c[N];
 7 __int64 count[N];
 8 int k;
 9 int cmp(const void *a,const void *b)
10 {
11     return *(int *)a - *(int *)b;
12 }
13 int find(__int64 x)
14 {
15     int left,right,mid;
16     left=1;
17     right=k;
18     while(right>=left)
19     {
20         mid=(left+right)/2;
21         if(b[mid]==x) return mid;
22         if(b[mid]>x) right=mid-1;
23         else left=mid+1;
24     }
25     return 0;
26 }
27 __int64 query(int i)
28 {
29     __int64 sum=0;
30     while(i>0)
31     {
32         sum+=count[i];
33         sum%=mod;
34         i-=i&(-i);
35     }
36     return sum;
37 }
38 void insert(int i,__int64 num)
39 {
40     while(i<=k)
41     {
42         count[i]+=num;
43         count[i]%=mod;
44         i+=i&(-i);
45     }
46 }
47 int main()
48 {
49     int i,m,ncase,t,ans,n;
50     __int64 sum,num,X,Y,Z;
51     scanf("%d",&ncase);
52     for(t=1;t<=ncase;t++)
53     {
54         scanf("%d%d%I64d%I64d%I64d",&n,&m,&X,&Y,&Z);
55         for(i=0;i<m;i++)
56             scanf("%d",&a[i]);
57         for(i=0;i<=n-1;i++)
58         {
59             b[i+1]=a[i%m];
60             c[i+1]=b[i+1];
61             a[i%m]=(X*a[i%m]+Y*(i+1))%Z;
62         }
63         qsort(b+1,n,sizeof(b[1]),cmp);
64         k=1;
65         for(i=2;i<=n;i++)
66             if(b[i]!=b[i-1]) b[++k]=b[i];
67         memset(count,0,sizeof(count));
68         sum=0;
69         for(i=1;i<=n;i++)
70         {
71             ans=find(c[i]);
72             num=query(ans-1);
73             num%=mod;
74             sum+=num+1;
75             sum%=mod;
76             insert(ans,num+1);
77         }
78         printf("Case #%d: %I64d\n",t,sum%mod);
79     }
80     return 0;
81 }