題目連結

題意：
給定一個長度為n$n$的十六進位制數，和m$m$次詢問，詢問格式為(op,x,y)$(op,x,y)$，當op=1$op=1$，表示將第x$x$個數位的數字改為y$y$，當op=2$op=2$時，問區間[x,y]$[x,y]$的價值為多少，其中區間[l,r]$[l,r]$區間的價值定義為：
取出[l,r]中的所有子串v$v$（不要求連續），計算SOD(v)$SOD(v)$，則區間價值為：
∑(numberofSOD(v)=i)∗1021i$\sum (numberofSOD(v)=i)\ast {1021}^i$ mod(1e9+7)$mod(1e9+7)$

而SOD(v)$SOD(v)$的定義為：

def SOD(v):
    if
 
 v < 16: 
        return v
    else:
        return SOD(sum of digit of v in hexadecimal)

思路：
首先對於SOD(v)$SOD(v)$，由同餘定理，可以得到其等價於：

def SOD(v):
    if v == 0: 
        return 0
    elif v % 15 == 0:
        return 15
    else:
        return v % 15

故對於區間價值的計算，i$i$只有16種取值。
故可以使用線段樹來維護區間中所有子序列的SOD值為i$i$

void push_up(int rt){
    for (int i = 0; i < 16; i++) {
        Tree[rt].sum[i] = (Tree[lson].sum[i] + Tree[rson].sum[i]) % mod;
    }
    for (int i = 0; i < 16; i++) {
        for (int j = 0; j < 16; j++) {
            int x = (i + j) % 15;
            if
 
 (!i && !j) x = 0;
            else if (x == 0) x = 15;
            Tree[rt].sum[x] = (Tree[rt].sum[x] + Tree[lson].sum[i] * Tree[rson].sum[j] % mod) % mod;
        }
    }
}

其他就是經典的線段樹操作了。
此題得解。

程式碼：

#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
#define lson rt<<1
#define rson rt<<1|1
const int mod = 1e9 + 7;
const int A = 1e5 + 10;
const int B = 20;
class Seg_Tree{
public:
    int l,r;
    ll sum[B];
}Tree[A<<2];
char s[A], t[B];
ll fac[B], res[B];
int n,q;

void Init(){
    fac[0] = 1;
    for (int i = 1; i < B; i++) {
        fac[i] = fac[i - 1] * 1021 % mod;
    }
}

int get_val(char c){
    if (c >= '0' && c <= '9') return c - '0';
    return c - 'A' + 10;
}

void push_up(int rt){
    for (int i = 0; i < 16; i++) {
        Tree[rt].sum[i] = (Tree[lson].sum[i] + Tree[rson].sum[i]) % mod;
    }
    for (int i = 0; i < 16; i++) {
        for (int j = 0; j < 16; j++) {
            int x = (i + j) % 15;
            if (!i && !j) x = 0;
            else if (x == 0) x = 15;
            Tree[rt].sum[x] = (Tree[rt].sum[x] + Tree[lson].sum[i] * Tree[rson].sum[j] % mod) % mod;
        }
    }
}

void build_Tree(int rt, int l, int r){
    Tree[rt].l = l, Tree[rt].r = r;
    for (int i = 0; i < 16; i++) Tree[rt].sum[i] = 0;
    if (l == r) {
        Tree[rt].sum[get_val(s[l])] = 1;
        return;
    }
    int mid = (l + r) >> 1;
    build_Tree(lson, l, mid);
    build_Tree(rson, mid + 1, r);
    push_up(rt);
}

void update(int rt, int pos, int c){
    int l = Tree[rt].l, r = Tree[rt].r;
    if (l == r) {
        for (int i = 0; i < 16; i++) Tree[rt].sum[i] = 0;
        Tree[rt].sum[c] = 1;
        return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid) update(lson, pos, c);
    else            update(rson, pos, c);
    push_up(rt);
}

void Merge(ll* now){
    ll tem[B] = {0};
    for (int i = 0; i < 16; i++) tem[i] = (res[i] + now[i]) % mod;
    for (int i = 0; i < 16; i++) {
        for (int j = 0; j < 16; j++) {
            int x = (i + j) % 15;
            if (!i && !j) x = 0;
            else if (x == 0) x = 15;
            tem[x] = (tem[x] + res[i] * now[j] % mod) % mod;
        }
    }
    for (int i = 0; i < 16; i++) res[i] = tem[i];
}

void query(int rt, int st, int ed){
    int l = Tree[rt].l, r = Tree[rt].r;
    if (st <= l && r <= ed) {
        Merge(Tree[rt].sum);
        return;
    }
    int mid = (l + r) >> 1;
    if (st <= mid) query(lson, st, ed);
    if (ed >  mid) query(rson, st, ed);
}

int main(){
    Init();
    scanf("%d%d", &n, &q);
    scanf("%s", s+1);
    build_Tree(1, 1, n);
    while (q--) {
        int op;
        scanf("%d", &op);
        if (op == 1) {
            int p;
            scanf("%d%s",&p,t);
            update(1, p, get_val(t[0]));
        } else {
            int l, r;
            scanf("%d%d", &l, &r);
            query(1, l, r);
            ll ans = 0;
            for (int i = 0; i < 16; i++) {
                ans = (ans + fac[i] * res[i] % mod) % mod;
                res[i] = 0;
            }
            printf("%lld\n", ans);
        }
    }
    return 0;
}