ZOJ3772-Calculate the Function

Calculate the Function

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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You are given a list of numbers A₁ A₂ .. A_N and M queries. For the i-th query:

• The query has two parameters L_i and R_i.
• The query will define a function F_i(x) on the domain [L_i, R_i] ∈ Z.
• F_i(L_i) = A_Li
• F_i(L_i + 1) = A_{(Li + 1)}
• for all x >= L_i + 2, F_i(x) = F_i(x - 1) + F_i(x - 2) × A_x

You task is to calculate F_i(R_i) for each query. Because the answer can be very large, you should output the remainder of the answer divided by 1000000007.

Input

There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:

The first line contains two integers N, M (1 <= N, M <= 100000). The second line contains N integers A₁ A₂ .. A_N (1 <= A_i <= 1000000000).

The next M lines, each line is a query with two integer parameters L_i, R_i (1 <= L_i <= R_i <= N).

Output

For each test case, output the remainder of the answer divided by 1000000007.

Sample Input

14 71 2 3 41 11 21 31 42 43 44 4

Sample Output

125131144

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Author: CHEN, Weijie
Source: The 14th Zhejiang University Programming Contest

题意：给一个序列An，有m个询问，每个询问包括l和r，定义f(l) = a[l], f(l+1) = a[l+1], f(x)=f(x-1) + a[x] * f(x-2), x >= l + 2;，对每个询问，求f(r)

解题思路：当x>=l+2时，f(x)=f(x-1) + a[x]* f(x-2), 所以就有递推式

所以当r>=l+1时，

用线段树来解决，每个节点保存一个矩阵

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>using namespace std;#define LL long longconst LL mod=1e9+7;LL a[100060];int n,m;struct node{    LL xx[3][3];    node(){memset(xx,0,sizeof xx);}}x[4*100060];node Union(node a,node b){    node ans;    for(int i=1; i<=2; i++)    {        for(int j=1; j<=2; j++)        {            for(int k=1; k<=2; k++)            {                ans.xx[i][j]+=a.xx[i][k]*b.xx[k][j];                ans.xx[i][j]%=mod;            }        }    }    return ans;}void build(int l,int r,int k){    if(l==r)    {        x[k].xx[1][1]=x[k].xx[1][2]=1;        x[k].xx[2][1]=a[r+1];        x[k].xx[2][2]=0;        return ;    }    int mid=(l+r)>>1;    build(l,mid,k<<1);    build(mid+1,r,k<<1|1);    x[k]=Union(x[k<<1],x[k<<1|1]);}node query(int l,int r,int ll,int rr,int k){    if(ll>=l&&rr<=r) return x[k];    int mid=(ll+rr)>>1;    node ans;    ans.xx[1][1]=ans.xx[2][2]=1;    if(l<=mid) ans=Union(ans,query(l,r,ll,mid,k<<1));    if(r>mid) ans=Union(ans,query(l,r,mid+1,rr,k<<1|1));    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++) scanf("%lld",&a[i]);        memset(x,0,sizeof x);        build(1,n-1,1);        while(m--)        {            int l,r;            scanf("%d%d",&l,&r);            if(l==r) printf("%lld\n",a[l]);            else if(r==l+1) printf("%lld\n",a[r]);            else            {                node ans=query(l+1,r-1,1,n-1,1);                printf("%lld\n",(a[l]*ans.xx[2][1]+a[l+1]*ans.xx[1][1])%mod);            }        }    }    return 0;}