进制(数据结构线段树)

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whiteawll 2022-08-16 14:53:47 博主文章分类：练习题目 ©著作权

文章标签 数据结构 p2p 网络协议 #define 进制 文章分类 数据结构与算法人工智能

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#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fir first
#define pb push_back
#define sec second
#define sortall(x) sort((x).begin(),(x).end())
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}

const int N = 4e5 + 10;
const int mod = 1e9 + 7;

string s;
struct bigtree {
    int l, r, ma;
}bt[N];
int n, q;


ll qmi (ll a, ll b) {
    ll ans = 1;
    while(b) {
        if (b & 1) ans =ans * a %mod;
        a = a * a %mod;
        b >>= 1;
    } 
    return ans;
}
void buildBT (int p, int l, int r) {
    bt[p].l = l, bt[p].r = r;
    if (l == r) {
        bt[p].ma = s[l - 1] - '0';
        return;
    }
    int mid = l + r>> 1;
    buildBT (p * 2, l, mid);
    buildBT (p * 2 + 1, mid+ 1, r);
    bt[p].ma = max (bt[p * 2].ma, bt[p * 2 + 1].ma);
}
int askma (int p, int l, int r) {
    int res =  0;
    if (bt[p].l >= l && bt[p].r <= r) {
        return bt[p].ma;
    }
    int mid = bt[p].l + bt[p].r >> 1;
    
    if (mid >= l) 
        res = max (res, askma(p* 2, l, r)); 
    if (mid + 1 <= r) 
        res = max (res, askma(p * 2 + 1, l, r));
    return res;
}
void changeBT (int p, int id, int x) {
    if (bt[p].l == id && bt[p].r == id) {
        bt[p].ma = x;
        return;
    }
    int mid = bt[p].l + bt[p].r >> 1;
    if (mid >= id) changeBT(p * 2, id, x);
    if (mid + 1 <= id) changeBT(p * 2 + 1, id, x);
    bt[p].ma = max (bt[p * 2].ma, bt[p * 2 + 1].ma); 
}

struct jztree {
    int l, r;
    ll res;
}t[N][11];

void buildJZ (int p, int l, int r, int x) {
    t[p][x].l = l, t[p][x].r = r;
    if (l == r) {
        t[p][x].res = s[l - 1] - '0';
        return;
    }
    int mid = l + r>> 1;
    buildJZ(p * 2, l, mid, x);
    buildJZ(p * 2 + 1, mid + 1, r, x);
    int len = t[p * 2 + 1][x].r - t[p * 2 + 1][x].l + 1;
    t[p][x].res = (qmi(x, len) * t[p * 2][x].res %mod + t[p * 2 + 1][x].res) %mod;
}
ll askans (int p, int l, int r, int x) {
    int res1 = 0, res2 = 0;
    if (t[p][x].l >= l && t[p][x].r <= r) {
        return t[p][x].res %mod;
    }
    int mid = t[p][x].l + t[p][x].r >> 1;
    if (mid >= l) {
        res1 = askans(p * 2, l , r, x);
    }
    if (mid + 1 <= r) 
    {
        res2 = askans(p * 2 + 1, l, r, x);
        int len = min (r, t[p * 2  + 1][x].r) - mid;
        return (res1 * qmi(x, len) %mod + res2) % mod;
    }
    else return res1 %mod;
}
void addans (int p, int id, int k, int x) {
    if (t[p][x].l == id && t[p][x].r == id) {
        t[p][x].res = k;
        return;
    }
    int mid = t[p][x].l + t[p][x].r >> 1;
    if (mid >= id)
        addans(p * 2, id, k, x);
    if (mid + 1 <= id)
        addans (p * 2 + 1, id, k, x);
    int len = t[p *2 + 1][x].r - t[p * 2 + 1][x].l + 1;
    t[p][x].res = (qmi(x, len) * t[p * 2][x].res + t[p * 2 + 1][x].res) %mod; 
}
void solve(){
    cin >> n >> q;
    cin >> s;
    buildBT (1, 1, n);
    for (int i =2; i<= 10; i++)
        buildJZ(1, 1, n, i);
    while(q --) {
        int op;
        cin >> op;
        int x, y;
        cin >> x >> y;
        if (op == 1) {
            changeBT (1, x, y);
            for (int i =2; i<= 10; i++)
                addans(1, x, y, i);
        }
        else {
            int ma = askma(1, x, y) + 1; 
            if (ma == 1) ma ++;
            cout << askans(1, x, y, ma) << endl;
        }
    }
   
}
int main () {
    int t;
    t = 1; 
    while (t --) solve();
    return 0;
}