Codeforces Round #FF (Div. 2)__E. DZY Loves Fibonacci Numbers (CF447) 线段树

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。

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int MAXN = 3e5+;
 const int MOD = 1e9+;
 LL sum[MAXN << ], addv1[MAXN << ], addv2[MAXN << ];
 LL fib[MAXN], hhh[MAXN], fff[MAXN];
 void pre_solve() {
     fff[] = ;
     fff[] = ;
     fib[] = fib[] = ;
     hhh[] = ;
     for (int i = ; i < MAXN; i++) {
         fib[i] = (fib[i-] + fib[i-]) % MOD;
         fff[i] = (fff[i-] + fff[i-]) % MOD;
     }
     for (int i = ; i < MAXN; i++) {
         hhh[i] = (fib[i-] + hhh[i-]) % MOD;
     }
 }
 void push_up(int pos) {
     sum[pos] = (sum[pos<<] + sum[pos<<|]) % MOD;
 }
 void update_add(int l, int r, int pos, LL val1, LL val2) {
     addv1[pos] = (addv1[pos] + val1) % MOD;
     addv2[pos] = (addv2[pos] + val2) % MOD;
     int idx = (r - l + );
     sum[pos] = (sum[pos] + fib[idx] * val1 % MOD + hhh[idx] * val2 % MOD) % MOD;
 }
 void push_down(int l, int r, int pos) {
     int mid = (l + r) >> ;
     update_add(l, mid, pos<<, addv1[pos], addv2[pos]);
     update_add(mid+, r, pos<<|, (fff[mid+-l+]*addv1[pos]+fib[mid+-l]*addv2[pos])%MOD,
                (fff[mid+-l+]*addv1[pos]+fib[mid+-l+]*addv2[pos])%MOD);
     addv1[pos] = addv2[pos] = ;
 }
 void build (int l, int r, int pos) {
     addv1[pos] = addv2[pos] = ;
     if (l == r) {
         scanf ("%I64d", sum+pos);
         return ;
     }
     int mid = (l + r) >> ;
     build(l, mid, pos<<);
     build(mid+, r, pos<<|);
     push_up(pos);
 }
 void update (int l, int r, int pos, int ua, int ub, LL x1, LL x2) {
     if (ua <= l && ub >= r) {
         update_add(l, r, pos, x1, x2);
         return ;
     }
     push_down(l, r, pos);
     int mid = (l + r) >> ;
     if (ua <= mid) {
         update(l, mid, pos<<, ua, ub, x1, x2);
     }
     if (ub > mid) {
         if (ua <= mid) {
             int tmp = max(l, ua);
             LL t1 = (fff[mid+-tmp+]*x1+fib[mid+-tmp]*x2) % MOD;
             LL t2 = (fff[mid+-tmp+]*x1+fib[mid+-tmp+]*x2) % MOD;
             update(mid+, r, pos<<|, ua, ub, t1, t2);
         } else {
             update(mid+, r, pos<<|, ua, ub, x1, x2);
         }
     }
     push_up(pos);
 }
 LL query (int l, int r, int pos, int ua, int ub) {
     if (ua <= l && ub >= r) {
         return sum[pos];
     }
     push_down(l, r, pos);
     int mid = (l + r) >> ;
     LL tmp = ;
     if (ua <= mid)
     {
         tmp = (tmp + query(l, mid, pos<<, ua, ub)) % MOD;
     }
     if (ub > mid)
     {
         tmp = (tmp + query(mid+, r, pos<<|, ua, ub)) % MOD;
     }

     return tmp;
 }
 int main() {
 #ifndef ONLINE_JUDGE
     freopen("in.txt","r",stdin);
 #endif
     int n, m;
     pre_solve();
     while (~ scanf ("%d%d", &n, &m)) {
         build(, n, );
         for (int i = ; i < m; i++) {
             int op, u, v;
             scanf ("%d%d%d", &op, &u, &v);
             if (op == ) {
                 update(, n, , u, v, fib[], fib[]);
             } else {
                 LL ans = query(, n, , u, v);
                 printf("%I64d\n", ans);
             }
         }
     }
     return ;
 }