題意在這裡就不說了。直接說思路，這是一道裸的線段樹，每個節點儲存區間的和sum以及區間變成斐波那契數的和fs。然後就是簡單點修改和區間修改的事了。

詳細見程式碼

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define N 100006
#define mid (l+r>>1)
#define lc (d<<1)
#define rc (d<<1|1)
#define ll long long
 

ll fb[1005];

struct Tr{
    ll sum, fs;
    int lz;
}tr[N<<2];

void Push(int d) {
    tr[d].sum = tr[lc].sum+tr[rc].sum;
    tr[d].fs = tr[lc].fs+tr[rc].fs;
}

void build(int d, int l, int r) {
    if (l == r) {
        tr[d].sum = 0, tr[d].fs = 1;
        tr[d].lz = 0;
        return;
    }
    tr[
 
d].lz = 0;
    build(lc, l, mid);
    build(rc, mid+1, r);
    Push(d);
}

void lazy(int d, int l, int r) {
    if (tr[d].lz) {
        tr[lc].lz = 1, tr[lc].sum = tr[lc].fs;
        tr[rc].lz = 1, tr[rc].sum = tr[rc].fs;
        tr[d].lz = 0;
    }
}

ll cal(ll a) {
    int f = 0, r = 90, m;
    while (
 
f < r) {
        m = (f+r)/2;
        //cout << f << "  "  << m << "  " << r << " " << fb[m] << endl;
        if (fb[m] < a) f = m+1;
        else r = m;
    }
   // printf("[%lld %d %lld]\n", a, f, fb[f]);
    if (f == 0) return fb[f];
    ll tm = fb[f]-a;
    if (tm < a-fb[f-1]) return fb[f];
    return fb[f-1];
}

void add(int d, int l, int r, int pos, int ad) {
    if (l == r && l == pos) {
        //if(tr[d].lz == 1) tr[d].sum = cal(tr[d].sum);
        tr[d].sum += ad; tr[d].fs = cal(tr[d].sum);
        tr[d].lz = 0;
        return;
    }
    lazy(d, l, r);
    if (pos <= mid) add(lc, l, mid, pos, ad);
    else add(rc, mid+1, r, pos, ad);
    Push(d);
}

ll query(int d, int l, int r, int L, int R) {
    if (l == L && r == R) {
        return tr[d].sum;
    }
    lazy(d, l, r);
    if (R <= mid) return query(lc, l, mid, L, R);
    else if (L > mid) return query(rc, mid+1, r, L, R);
    else return query(lc, l, mid, L, mid)+query(rc, mid+1, r, mid+1, R);
}

void update(int d, int l, int r, int L, int R) {
    if (l == L && r == R) {
        tr[d].sum = tr[d].fs;
        tr[d].lz = 1;
        return;
    }
    lazy(d, l, r);
    if (R <= mid) update(lc, l, mid, L, R);
    else if (L > mid) update(rc, mid+1, r, L, R);
    else update(lc, l, mid, L, mid), update(rc, mid+1, r, mid+1, R);
    Push(d);
}

int main() {
    int n, m, i, j, q;
    memset(fb, 0, sizeof(fb));
    fb[0] = 1, fb[1] = 2;
    for (i = 2;i < 91;i++) fb[i] = fb[i-1]+(ll)fb[i-2];

  //  for (i = 0;i < 91;i++) cout << fb[i] << endl;
 // cout<<cal(1000000000)<<endl;
    while (~scanf("%d%d", &n, &m)) {
        build(1, 1, n);
        while (m--) {
            scanf("%d", &q);
            scanf("%d%d", &i, &j);
            if (q == 1) {
                add(1, 1, n, i, j);
            }else if (q == 2) {
                cout << query(1, 1, n, i, j) << endl;
            }else {
                update(1, 1, n, i, j);
            }
        }
    }
    return 0;
}