You are given a list of numbers A1 A2 .. AN and M queries. For the i-th query:
You task is to calculate Fi(Ri) for each query. Because the answer can be very large, you should output the remainder of the answer divided by 1000000007.
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 .. AN (1 <= Ai <= 1000000000).
The next M lines, each line is a query with two integer parameters Li, Ri (1 <= Li <= Ri <= N).
For each test case, output the remainder of the answer divided by 1000000007.
1 4 7 1 2 3 4 1 1 1 2 1 3 1 4 2 4 3 4 4 4
1 2 5 13 11 4 4
Author: CHEN, Weijie
Source: The 14th Zhejiang
University Programming Contest
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<cstdlib> 5 using namespace std; 6 7 typedef long long LL; 8 9 int mod=1000000007; 10 int ax[100002]; 11 struct node 12 { 13 LL a,b,c,d; 14 int l,r; 15 } f[400008]; 16 17 void build(int l,int r,int n) 18 { 19 int mid=(l+r)/2; 20 f[n].l=l; 21 f[n].r=r; 22 if(l==r) 23 { 24 f[n].a=0; 25 f[n].b=ax[l]; 26 f[n].c=1; 27 f[n].d=1; 28 return; 29 } 30 build(l,mid,n*2); 31 build(mid+1,r,n*2+1); 32 f[n].a=((f[n<<1].a*f[(n<<1)+1].a)%mod+f[n<<1].b*f[(n<<1)+1].c)%mod; 33 f[n].b=((f[n<<1].a*f[(n<<1)+1].b)%mod+f[n<<1].b*f[(n<<1)+1].d)%mod; 34 f[n].c=((f[n<<1].c*f[(n<<1)+1].a)%mod+f[n<<1].d*f[(n<<1)+1].c)%mod; 35 f[n].d=((f[n<<1].c*f[(n<<1)+1].b)%mod+f[n<<1].d*f[(n<<1)+1].d)%mod; 36 } 37 node serch1(int l,int r,int n) 38 { 39 int mid=(f[n].l+f[n].r)/2; 40 node n1,n2,n3; 41 42 if(f[n].l==l && f[n].r==r) 43 { 44 return f[n]; 45 } 46 if(mid>=r) 47 return serch1(l,r,n*2); 48 else if(mid<l) 49 return serch1(l,r,n*2+1); 50 else 51 { 52 n1=serch1(l,mid,n*2); 53 n2=serch1(mid+1,r,n*2+1); 54 n3.a=((n1.a*n2.a)%mod+(n1.b*n2.c)%mod)%mod; 55 n3.b=((n1.a*n2.b)%mod+(n1.b*n2.d)%mod)%mod; 56 n3.c=((n1.c*n2.a)%mod+(n1.d*n2.c)%mod)%mod; 57 n3.d=((n1.c*n2.b)%mod+(n1.d*n2.d)%mod)%mod; 58 } 59 return n3; 60 } 61 int main() 62 { 63 int T; 64 int i,j,n,m,x,y; 65 LL sum1; 66 node cur; 67 scanf("%d",&T); 68 while(T--) 69 { 70 scanf("%d%d",&n,&m); 71 for(i=1; i<=n; i++) 72 scanf("%d",&ax[i]); 73 build(1,n,1); 74 for(j=1; j<=m; j++) 75 { 76 scanf("%d%d",&x,&y); 77 if(y-x<2) 78 { 79 printf("%d\n",ax[y]); 80 } 81 else 82 { 83 cur=serch1(x+2,y,1); 84 sum1=((cur.b*ax[x])%mod+(cur.d*ax[x+1])%mod)%mod; 85 printf("%lld\n",sum1); 86 } 87 } 88 } 89 return 0; 90 }
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原文：http://www.cnblogs.com/tom987690183/p/3651216.html