H. Fibonacci-ish II
題目連接：
http://codeforces.com/contest/633/problem/H
Description
Yash is finally tired of computing the length of the longest Fibonacci-ish sequence. He now plays around with more complex things such as Fibonacci-ish potentials.
Fibonacci-ish potential of an array ai is computed as follows:
Remove all elements j if there exists i < j such that ai = aj.
Sort the remaining elements in ascending order, i.e. a1 < a2 < ... < an.
Compute the potential as P(a) = a1·F1 + a2·F2 + ... + an·Fn, where Fi is the i-th Fibonacci number (see notes for clarification).
You are given an array ai of length n and q ranges from lj to rj. For each range j you have to compute the Fibonacci-ish potential of the array bi, composed using all elements of ai from lj to rj inclusive. Find these potentials modulo m.
Input
The first line of the input contains integers of n and m (1 ≤ n, m ≤ 30 000) — the length of the initial array and the modulo, respectively.
The next line contains n integers ai (0 ≤ ai ≤ 109) — elements of the array.
Then follow the number of ranges q (1 ≤ q ≤ 30 000).
Last q lines contain pairs of indices li and ri (1 ≤ li ≤ ri ≤ n) — ranges to compute Fibonacci-ish potentials.
Output
Print q lines, i-th of them must contain the Fibonacci-ish potential of the i-th range modulo m.
Sample Input
5 10
2 1 2 1 2
2
2 4
4 5
Sample Output
3
3
Hint
題意
給你n個數，然后有q次詢問
每次詢問給你l,r區間
你首先得把l,r區間的數全部拿出來，從小到大排序，然后再去重
然后答案就等於ans = f[1]*a[1]+f[2]*a[2]....+f[n]*a[n]
其中f[i]是第i個fib數
要求答案mod m
題解：
現在出題人總喜歡艹一些大新聞出來
出一些復雜度迷的題目
這道題正解是莫隊+奇怪的東西去維護
但是由於常數太大了，還沒有大力出奇跡的暴力跑得快……
你說腫么辦？
正解如下：
莫隊去處理詢問，然后線段樹去處理答案
線段樹怎么處理呢？
這里比較麻煩處理的是push_up
push_up你可以用矩陣去寫。
每個點維護兩個矩陣,[1][2]是答案的矩陣，[2][2]是fib的矩陣
a[i][1][2] = a[i*2][1][2] + a[i*2+1][1][2] * a[i*2][2][2]
a[i][2][2] = a[i*2][2][2]*a[i*2+1][2][2]
乘號是矩陣乘法
當然，用數學方法也可以
f(k+1)*(a1*f(i)+a2*f(i+1)+...)+f(k)*(a1*(f(i-1))+a2*f(i)....) = a1*f(i+k)+a2*f(i+k+1).....
大暴力代碼
#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e4+5;
pair<int,int> a[maxn];
int ans[maxn],step[maxn],f[maxn],l[maxn],r[maxn],last[maxn];
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&a[i].first),a[i].second=i;
sort(a+1,a+1+n);
f[0]=1,f[1]=1;
for(int i=2;i<=n;i++)
f[i]=(f[i-1]+f[i-2])%m;
int q;scanf("%d",&q);
for(int i=1;i<=q;i++)
{
scanf("%d%d",&l[i],&r[i]);
last[i]=-1;
}
for(int i=1;i<=n;i++)
{
int d = a[i].first % m;
for(int j=1;j<=q;j++)
{
if(a[i].second<l[j]||a[i].second>r[j])continue;
if(a[i].first==last[j])continue;
ans[j]=(ans[j]+f[step[j]++]*d)%m;
last[j]=a[i].first;
}
}
for(int i=1;i<=q;i++)
printf("%d\n",ans[i]);
}
莫隊
#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e4+5;
int a[maxn],pos[maxn],q,n,m,d[maxn];
int fib[maxn][2];
map<int,int> H;
vector<int> V;
long long Ans[maxn];
int cnt[maxn];
typedef int SgTreeDataType;
struct treenode
{
int L , R ;
SgTreeDataType sum1,sum2,siz;
void update(SgTreeDataType v)
{
siz = v;
sum1 = sum2 = v*d[L]%m;
}
};
treenode tree[maxn*4];
inline void push_down(int o)
{
}
inline void push_up(int o)
{
tree[o].siz = tree[2*o].siz + tree[2*o+1].siz;
tree[o].sum1 = (tree[2*o].sum1+fib[tree[2*o].siz][0]*tree[o*2+1].sum1+fib[tree[2*o].siz][1]*tree[o*2+1].sum2)%m;
tree[o].sum2 = (tree[2*o].sum2+fib[tree[2*o].siz+1][0]*tree[o*2+1].sum1+fib[tree[2*o].siz+1][1]*tree[o*2+1].sum2)%m;
}
inline void build_tree(int L , int R , int o)
{
tree[o].L = L , tree[o].R = R,tree[o].sum1 = tree[o].sum2 = tree[o].siz = 0;
if (R > L)
{
int mid = (L+R) >> 1;
build_tree(L,mid,o*2);
build_tree(mid+1,R,o*2+1);
}
}
inline void update(int QL,int QR,SgTreeDataType v,int o)
{
int L = tree[o].L , R = tree[o].R;
if (QL <= L && R <= QR) tree[o].update(v);
else
{
push_down(o);
int mid = (L+R)>>1;
if (QL <= mid) update(QL,QR,v,o*2);
if (QR > mid) update(QL,QR,v,o*2+1);
push_up(o);
}
}
struct query
{
int l,r,id;
}Q[maxn];
bool cmp(query a,query b)
{
if(pos[a.l]==pos[b.l])
return a.r<b.r;
return pos[a.l]<pos[b.l];
}
void Updata(int x)
{
int p = H[a[x]];
cnt[p]++;
if(cnt[p]==1)
update(p,p,1,1);
}
void Delete(int x)
{
int p = H[a[x]];
cnt[p]--;
if(cnt[p]==0)
update(p,p,0,1);
}
int main()
{
scanf("%d%d",&n,&m);
int sz =ceil(sqrt(1.0*n));
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
V.push_back(a[i]);
pos[i]=(i-1)/sz;
}
sort(V.begin(),V.end());
V.erase(unique(V.begin(),V.end()),V.end());
for(int i=0;i<V.size();i++)
{
H[V[i]]=i+1;
d[i+1]=V[i]%m;
}
fib[0][0]=1%m,fib[1][1]=1%m;
for(int i=2;i<=n;i++)
{
fib[i][0]=(fib[i-1][0]+fib[i-2][0])%m;
fib[i][1]=(fib[i-1][1]+fib[i-2][1])%m;
}
build_tree(1,n,1);
scanf("%d",&q);
for(int i=1;i<=q;i++)
{
scanf("%d%d",&Q[i].l,&Q[i].r);
Q[i].id = i;
}
sort(Q+1,Q+1+q,cmp);
int l = 1,r = 0;
int ans=0;
for(int i=1;i<=q;i++)
{
int id = Q[i].id;
while(r<Q[i].r)
{
r++;
Updata(r);
}
while(l>Q[i].l)
{
l--;
Updata(l);
}
while(r>Q[i].r)
{
Delete(r);
r--;
}
while(l<Q[i].l)
{
Delete(l);
l++;
}
Ans[id]=tree[1].sum1;
}
for(int i=1;i<=q;i++)
printf("%lld\n",Ans[i]);
}