zoj 3772 Calculate the Function

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5235

这道题需要构造矩阵：F(X)=F(X-1)+F(X-2)*A(X)转化为F(X)*A(X+2)+F(X+1)=F(X+2)，然后在构造矩阵

{1, A[x]}  {F(x+1)}  {F(X+2)}

{1,    0 }*{F(X)    }={F(X+1)}

然后用线段数维护乘号左边的乘积；

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #define maxn 101000
 using namespace std;
 const int mod=;

 int t,a[maxn*],n,m;
 int l,r;
 struct matrix
 {
     long long a[][];
 };

 struct node
 {
     int l;
     int r;
     matrix c;
 } tree[maxn*];

 matrix multi(matrix x,matrix y)
 {
     matrix temp;
     for(int i=; i<=; i++)
     {
         for(int j=; j<=; j++)
         {
             temp.a[i][j]=;
             for(int k=; k<=; k++)
             {
                 temp.a[i][j]=(x.a[i][k]*y.a[k][j]+temp.a[i][j])%mod;
             }
         }
     }
     return temp;
 }

 matrix build(int i,int l,int r)
 {
     tree[i].l=l;
     tree[i].r=r;
     if(l==r)
     {
         tree[i].c.a[][]=;
         tree[i].c.a[][]=a[l];
         tree[i].c.a[][]=;
         tree[i].c.a[][]=;
         return tree[i].c;
     }
     int mid=(l+r)>>;
     build(i<<,l,mid);
     build(i<<|,mid+,r);
     tree[i].c=multi(tree[i<<|].c,tree[i<<].c);
     return tree[i].c;
 }

 matrix search1(int i,int l,int r)
 {
     if(tree[i].l==l&&tree[i].r==r)
     {
         return tree[i].c;
     }
     int mid=(tree[i].l+tree[i].r)>>;
     if(r<=mid)
     {
         return search1(i<<,l,r);
     }
     else if(l>mid)
     {
         return search1(i<<|,l,r);
     }
     else
     {
         return multi(search1(i<<|,mid+,r),search1(i<<,l,mid));
     }
 }
 int main()
 {
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&m);
         for(int i=; i<=n; i++)
         {
             scanf("%d",&a[i]);
         }
         build(,,n);
         while(m--)
         {
             scanf("%d%d",&l,&r);
             if(r-l>=)
             {
                 matrix tm,ans;
                 tm.a[][]=a[l+];
                 tm.a[][]=a[l];
                 ans=multi(search1(,l+,r),tm);
                 printf("%lld\n",ans.a[][]);
             }
             else printf("%d\n",a[r]%mod);
         }
     }
     return ;
 }