题目链接：hdu 4983 Wow! Such Sequence!

题目大意：就是三种操作

1 k d， 改动k的为值添加d

2 l r， 查询l到r的区间和

3 l r。 间l到r区间上的所以数变成近期的斐波那契数，相等的话取向下取。

解题思路：线段树。对于每一个节点新增一个bool表示该节点下面的位置是否都是斐波那契数。

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <algorithm>

#define lson(x) ((x<<1))

#define rson(x) ((x<<1)|1)

using namespace std;

typedef __int64 ll;

const int maxn = 100005;

const int maxf = 100;

const ll INF = 2000000000000000LL;

int n, m;

ll fib[maxf], fn, num[maxn];

struct maxnode {

    int l, r;

    ll sum;

    bool isfib;

    void set (int l, int r, ll sum, bool isfib) {

        this->l = l;

        this->r = r;

        this->sum = sum;

        this->isfib = isfib;

    }

} node[4 * maxn];

void init () {

    fib[0] = fib[1] = 1;

    for (fn = 2;; fn++) {

        fib[fn] = fib[fn - 2] + fib[fn - 1];

        if (fib[fn] > INF)

            break;

    }

}

void pushup(int x) {

    int l = lson(x), r = rson(x);

    node[x].isfib = (node[l].isfib && node[r].isfib);

    node[x].sum = node[l].sum + node[r].sum;;

}

void build(int l, int r, int x) {

    node[x].set(l, r, 0, false);

    if (l == r)

        return;

    int mid = (l + r) / 2;

    build(l, mid, lson(x));

    build(mid + 1, r, rson(x));

}

ll find (ll x) {

    int id;

    ll ans = INF;

    for (int i = 0; i < fn; i++) {

        ll k = (fib[i] > x ?

fib[i] - x : x - fib[i]);

        if (k < ans) {

            ans = k;

            id = i;

        }

    }

    return fib[id];

}

void add (int k, ll v, int x) {

    if (node[x].l == k && node[x].r == k) {

        node[x].sum += v;

        node[x].isfib = (find(node[x].sum) == node[x].sum ? true : false);

        return;

    }

    int mid = (node[x].l + node[x].r) / 2;

    if (k <= mid)

        add(k, v, lson(x));

    else if (k > mid)

        add(k, v, rson(x));

    pushup(x);

}

void insert(int l, int r, int x) {

    if (node[x].isfib)

        return;

    if (node[x].l == node[x].r) {

        node[x].sum = find(node[x].sum);

        node[x].isfib = true;

        return;

    }

    int mid = (node[x].l + node[x].r) / 2;

    if (l <= mid)

        insert(l, r, lson(x));

    if (r > mid)

        insert(l, r, rson(x));

    pushup(x);

}

ll query(int l, int r, int x) {

    if (node[x].l >= l && node[x].r <= r)

        return node[x].sum;

    int mid = (node[x].l + node[x].r) / 2;

    ll ans = 0;

    if (l <= mid)

        ans += query(l, r, lson(x));

    if (r > mid)

        ans += query(l, r, rson(x));

    return ans;

}

int main() {

    init();

    while (scanf("%d%d", &n, &m) == 2) {

        build(1, n, 1);

        int Q, a, b;

        ll v;

        while (m--) {

            scanf("%d", &Q);

            if (Q == 1) {

                scanf("%d%I64d", &a, &v);

                add(a, v, 1);

            } else if (Q == 2) {

                scanf("%d%d", &a, &b);

                printf("%I64d\n", query(a, b, 1));

            } else {

                scanf("%d%d", &a, &b);

                insert(a, b, 1);

            }

        }

    }

    return 0;

}