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；
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define maxn 101000 5 using namespace std; 6 const int mod=1000000007; 7 8 int t,a[maxn*10],n,m; 9 int l,r; 10 struct matrix 11 { 12 long long a[4][4]; 13 }; 14 15 struct node 16 { 17 int l; 18 int r; 19 matrix c; 20 } tree[maxn*4]; 21 22 matrix multi(matrix x,matrix y) 23 { 24 matrix temp; 25 for(int i=1; i<=2; i++) 26 { 27 for(int j=1; j<=2; j++) 28 { 29 temp.a[i][j]=0; 30 for(int k=1; k<=2; k++) 31 { 32 temp.a[i][j]=(x.a[i][k]*y.a[k][j]+temp.a[i][j])%mod; 33 } 34 } 35 } 36 return temp; 37 } 38 39 matrix build(int i,int l,int r) 40 { 41 tree[i].l=l; 42 tree[i].r=r; 43 if(l==r) 44 { 45 tree[i].c.a[1][1]=1; 46 tree[i].c.a[1][2]=a[l]; 47 tree[i].c.a[2][1]=1; 48 tree[i].c.a[2][2]=0; 49 return tree[i].c; 50 } 51 int mid=(l+r)>>1; 52 build(i<<1,l,mid); 53 build(i<<1|1,mid+1,r); 54 tree[i].c=multi(tree[i<<1|1].c,tree[i<<1].c); 55 return tree[i].c; 56 } 57 58 matrix search1(int i,int l,int r) 59 { 60 if(tree[i].l==l&&tree[i].r==r) 61 { 62 return tree[i].c; 63 } 64 int mid=(tree[i].l+tree[i].r)>>1; 65 if(r<=mid) 66 { 67 return search1(i<<1,l,r); 68 } 69 else if(l>mid) 70 { 71 return search1(i<<1|1,l,r); 72 } 73 else 74 { 75 return multi(search1(i<<1|1,mid+1,r),search1(i<<1,l,mid)); 76 } 77 } 78 int main() 79 { 80 scanf("%d",&t); 81 while(t--) 82 { 83 scanf("%d%d",&n,&m); 84 for(int i=1; i<=n; i++) 85 { 86 scanf("%d",&a[i]); 87 } 88 build(1,1,n); 89 while(m--) 90 { 91 scanf("%d%d",&l,&r); 92 if(r-l>=2) 93 { 94 matrix tm,ans; 95 tm.a[1][1]=a[l+1]; 96 tm.a[2][1]=a[l]; 97 ans=multi(search1(1,l+2,r),tm); 98 printf("%lld ",ans.a[1][1]); 99 } 100 else printf("%d ",a[r]%mod); 101 } 102 } 103 return 0; 104 }