http://codeforces.com/contest/447/problem/E
题意： 给定一个数组， m次操作，
1 l r 表示区间修改， 每次 a[i] + Fibonacci[i-l+]
2 l r 区间求和
每次修改操作，只需要记录 每个点前两个值就可以了， 后面的和以及孩子需要加的值都可以通过这两个求出来。 推推公式就出来了。
注意 溢出。
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int MAXN = 3e5+10; 5 const int MOD = 1e9+9; 6 LL sum[MAXN << 2], addv1[MAXN << 2], addv2[MAXN << 2]; 7 LL fib[MAXN], hhh[MAXN], fff[MAXN]; 8 void pre_solve() { 9 fff[1] = 1; 10 fff[2] = 0; 11 fib[1] = fib[2] = 1; 12 hhh[1] = 0; 13 for (int i = 3; i < MAXN; i++) { 14 fib[i] = (fib[i-1] + fib[i-2]) % MOD; 15 fff[i] = (fff[i-1] + fff[i-2]) % MOD; 16 } 17 for (int i = 2; i < MAXN; i++) { 18 hhh[i] = (fib[i-1] + hhh[i-1]) % MOD; 19 } 20 } 21 void push_up(int pos) { 22 sum[pos] = (sum[pos<<1] + sum[pos<<1|1]) % MOD; 23 } 24 void update_add(int l, int r, int pos, LL val1, LL val2) { 25 addv1[pos] = (addv1[pos] + val1) % MOD; 26 addv2[pos] = (addv2[pos] + val2) % MOD; 27 int idx = (r - l + 1); 28 sum[pos] = (sum[pos] + fib[idx] * val1 % MOD + hhh[idx] * val2 % MOD) % MOD; 29 } 30 void push_down(int l, int r, int pos) { 31 int mid = (l + r) >> 1; 32 update_add(l, mid, pos<<1, addv1[pos], addv2[pos]); 33 update_add(mid+1, r, pos<<1|1, (fff[mid+1-l+1]*addv1[pos]+fib[mid+1-l]*addv2[pos])%MOD, 34 (fff[mid+1-l+2]*addv1[pos]+fib[mid+1-l+1]*addv2[pos])%MOD); 35 addv1[pos] = addv2[pos] = 0; 36 } 37 void build (int l, int r, int pos) { 38 addv1[pos] = addv2[pos] = 0; 39 if (l == r) { 40 scanf ("%I64d", sum+pos); 41 return ; 42 } 43 int mid = (l + r) >> 1; 44 build(l, mid, pos<<1); 45 build(mid+1, r, pos<<1|1); 46 push_up(pos); 47 } 48 void update (int l, int r, int pos, int ua, int ub, LL x1, LL x2) { 49 if (ua <= l && ub >= r) { 50 update_add(l, r, pos, x1, x2); 51 return ; 52 } 53 push_down(l, r, pos); 54 int mid = (l + r) >> 1; 55 if (ua <= mid) { 56 update(l, mid, pos<<1, ua, ub, x1, x2); 57 } 58 if (ub > mid) { 59 if (ua <= mid) { 60 int tmp = max(l, ua); 61 LL t1 = (fff[mid+1-tmp+1]*x1+fib[mid+1-tmp]*x2) % MOD; 62 LL t2 = (fff[mid+1-tmp+2]*x1+fib[mid+1-tmp+1]*x2) % MOD; 63 update(mid+1, r, pos<<1|1, ua, ub, t1, t2); 64 } else { 65 update(mid+1, r, pos<<1|1, ua, ub, x1, x2); 66 } 67 } 68 push_up(pos); 69 } 70 LL query (int l, int r, int pos, int ua, int ub) { 71 if (ua <= l && ub >= r) { 72 return sum[pos]; 73 } 74 push_down(l, r, pos); 75 int mid = (l + r) >> 1; 76 LL tmp = 0; 77 if (ua <= mid) 78 { 79 tmp = (tmp + query(l, mid, pos<<1, ua, ub)) % MOD; 80 } 81 if (ub > mid) 82 { 83 tmp = (tmp + query(mid+1, r, pos<<1|1, ua, ub)) % MOD; 84 } 85 86 return tmp; 87 } 88 int main() { 89 #ifndef ONLINE_JUDGE 90 freopen("in.txt","r",stdin); 91 #endif 92 int n, m; 93 pre_solve(); 94 while (~ scanf ("%d%d", &n, &m)) { 95 build(1, n, 1); 96 for (int i = 0; i < m; i++) { 97 int op, u, v; 98 scanf ("%d%d%d", &op, &u, &v); 99 if (op == 1) { 100 update(1, n, 1, u, v, fib[1], fib[2]); 101 } else { 102 LL ans = query(1, n, 1, u, v); 103 printf("%I64d ", ans); 104 } 105 } 106 } 107 return 0; 108 }