ZOJ 3772 Calculate the Function

矩阵+线段树。。。

Fx            1  Ar                            1   A l+2          F l

Fx-1   =    1   0     *     ....    *      1    0            *   F l+1

注意矩阵乘的顺序。。。。

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

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long int LL;const int maxn=200000;const LL MOD=1000000007;struct MARTRIX{    LL m[2][2];    MARTRIX operator*(const MARTRIX& b) const    {        MARTRIX ret;        ret.m[0][0]=((m[0][0]*b.m[0][0])%MOD+(m[0][1]*b.m[1][0])%MOD)%MOD;        ret.m[0][1]=((m[0][0]*b.m[0][1])%MOD+(m[0][1]*b.m[1][1])%MOD)%MOD;        ret.m[1][0]=((m[1][0]*b.m[0][0])%MOD+(m[1][1]*b.m[1][0])%MOD)%MOD;        ret.m[1][1]=((m[1][0]*b.m[0][1])%MOD+(m[1][1]*b.m[1][1])%MOD)%MOD;        return ret;    }    MARTRIX(LL a)    {        m[0][0]=1; m[0][1]=a;        m[1][0]=1; m[1][1]=0;    }    MARTRIX()    {        memset(m,0,sizeof(m));    }}mrt[maxn<<2];int n,m;LL a[maxn];void build(int l,int r,int rt){    if(l==r)    {        mrt[rt]=MARTRIX(a[l]);        return ;    }    int m=(l+r)/2;    build(lson); build(rson);    mrt[rt]=mrt[rt<<1|1]*mrt[rt<<1];}MARTRIX query(int L,int R,int l,int r,int rt){    if(L<=l&&R>=r)    {        return mrt[rt];    }    int m=(l+r)/2;    if(R<=m)        return query(L,R,lson);    if(L>m)        return query(L,R,rson);    return query(L,R,rson)*query(L,R,lson);}int main(){    int T_T;    scanf("%d",&T_T);    while(T_T--)    {        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++) scanf("%lld",a+i);        build(1,n,1);        while(m--)        {            int L,R;            scanf("%d%d",&L,&R);            if(R<L+2)            {                printf("%lld\n",a[R]%MOD);                continue;            }            MARTRIX mt=query(L+2,R,1,n,1);            printf("%lld\n",((a[L+1]*mt.m[0][0])%MOD+(a[L]*mt.m[0][1])%MOD)%MOD);        }    }    return 0;}