H. Fibonacci-ish II
题目连接：
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
题意
给你n个数，然后有q次询问
每次询问给你l,r区间
你首先得把l,r区间的数全部拿出来，从小到大排序，然后再去重
然后答案就等于ans = f[1]*a[1]+f[2]*a[2]….+f[n]*a[n]
其中f[i]是第i个fib数
要求答案mod m
题解：
现在出题人总喜欢艹一些大新闻出来
出一些复杂度迷的题目
这道题正解是莫队+奇怪的东西去维护
但是由于常数太大了，还没有大力出奇迹的暴力跑得快……
你说肿么办？
正解如下：
莫队去处理询问，然后线段树去处理答案
线段树怎么处理呢？
这里比较麻烦处理的是push_up
push_up你可以用矩阵去写。
每个点维护两个矩阵,[1][2]是答案的矩阵，[2][2]是fib的矩阵
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]
乘号是矩阵乘法
当然，用数学方法也可以
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)…..
大暴力代码
#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]);
}
莫队
#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]);
}
