Bankopolis, the city you already know, finally got a new bank opened! Unfortunately, its security system is not yet working fine... Meanwhile hacker Leha arrived in Bankopolis and decided to test the system!

Bank has n cells for clients‘ money. A sequence from n

• 1 l r x y denoting that Leha changes each digit x to digit y in each element of sequence ai, for which l ≤ i ≤ r is holds. For example, if we change in number 11984381 digit 8 to 4, we get 11944341. It‘s worth noting that Leha, in order to stay in the shadow, never changes digits in the database to 0, i.e. y
   ≠ 0.
• 2 l r denoting that Leha asks to calculate and print the sum of such elements of sequence ai, for which l ≤ i ≤ r holds.

As Leha is a white-hat hacker, he don‘t want to test this vulnerability on a real database. You are to write a similar database for Leha to test.

The first line of input contains two integers n and q (1 ≤ n ≤ 105, 1 ≤ q ≤ 105) denoting amount of cells in the bank and total amount of queries respectively.

The following line contains n integers a1, a2, ..., an (1 ≤ ai < 109) denoting the amount of money in each cell initially. These integers do not contain leading zeros.

Each of the following q lines has one of the formats:

• 1 l r x y (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 9, 1 ≤ y ≤ 9), denoting Leha asks to change each digit x on digit y for each element ai of the sequence for which l ≤ i ≤ r holds;
• 2 l r (1 ≤ l ≤ r ≤ n), denoting you have to calculate and print the sum of elements ai for which l ≤ i ≤ r holds.

For each second type query print a single number denoting the required sum.

5 5
38 43 4 12 70
1 1 3 4 8
2 2 4
1 4 5 0 8
1 2 5 8 7
2 1 5

103
207

Let‘s look at the example testcase.

Initially the sequence is [38, 43, 4, 12, 70].

After the first change each digit equal to 4 becomes 8 for each element with index in interval [1; 3]. Thus, the new sequence is [38, 83, 8, 12, 70].

The answer for the first sum‘s query is the sum in the interval [2; 4], which equal 83 + 8 + 12 = 103, so the answer to this query is 103.

The sequence becomes [38, 83, 8, 12, 78] after the second change and [38, 73, 7, 12, 77] after the third.

The answer for the second sum‘s query is 38 + 73 + 7 + 12 + 77 = 207.

題意：

　　給你n個數

　　操作1：l r x y，區間[l,r]內所有數，數位上為x的都轉化為y

　　操作2: l r 求區間和

題解：

　　線段樹區間合並

　　建立10顆線段樹，分別表示數字0~9所代表的值

　　將x轉化為y也就是在將第x顆線段樹區間[l,r]和減去，加到第y顆線段樹上

　　這裏的延時操作有點小技巧

　　每次push_down的時候保持每個點(0~9)指向唯一的另外一個點，這樣再更新的時候才不會超時

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 5e5+10, M = 1e3+20, mod = 1e9+7,inf = 2e9;

LL sum[N][11],H[N],a[N],sum2[12];
int lazy[N][11],vis[12];

void push_down(int i,int ll,int rr) {
    if(ll == rr) return ;

    for(int j = 0; j < 10; ++j) vis[j] = lazy[ls][j], sum2[j] = sum[ls][j];
    for(int j = 0; j < 10; ++j) {
        if(lazy[i][j] != j) {
            for(int k = 0; k < 10; ++k) {
                if(lazy[ls][k] == j) vis[k] = lazy[i][j];
            }
            sum2[lazy[i][j]] += sum[ls][j]; sum2[j] -= sum[ls][j];
        }
    }
    for(int j = 0; j < 10; ++j)
        lazy[ls][j] = vis[j], sum[ls][j] = sum2[j];

    for(int j = 0; j < 10; ++j) vis[j] = lazy[rs][j], sum2[j] = sum[rs][j];
    for(int j = 0; j < 10; ++j) {
        if(lazy[i][j] != j) {
            for(int k = 0; k < 10; ++k) {
                if(lazy[rs][k] == j) vis[k] = lazy[i][j];
            }sum2[lazy[i][j]] += sum[rs][j]; sum2[j] -= sum[rs][j];
        }
    }
    for(int j = 0; j < 10; ++j)
        lazy[rs][j] = vis[j], sum[rs][j] = sum2[j];

    for(int j = 0; j < 10; ++j) lazy[i][j] = j;
}

void push_up(int i,int ll,int rr) {
    for(int j = 0; j <= 9; ++j) {
        sum[i][j] = sum[ls][j] + sum[rs][j];
    }
}

void build(int i,int ll,int rr) {

    for(int j = 0; j < 10; ++j) lazy[i][j] = j;

    if(ll == rr) {
        for(int j = 0; j < 10; ++j) sum[i][j]  =0;
        LL tmp = a[ll];
        for(int j = 1; j <= 12; ++j) {
            sum[i][tmp%10] += H[j-1];
            tmp/=10;
            if(tmp == 0) break;
        }
        return ;
    }

    build(ls,ll,mid); build(rs,mid+1,rr);
    push_up(i,ll,rr);
}

void update(int i,int ll,int rr,int x,int y,int f,int s) {

    push_down(i,ll,rr);
    if(ll == x && rr == y) {
        for(int j = 0; j <= 9; ++j)
        if(lazy[i][j] == f) {
            lazy[i][j] = s;
            sum[i][s] += sum[i][f];
            sum[i][f] = 0;
        }
        return ;
    }
    if(y <= mid) update(ls,ll,mid,x,y,f,s);
    else if(x > mid) update(rs,mid+1,rr,x,y,f,s);
    else {

        update(ls,ll,mid,x,mid,f,s);
        update(rs,mid+1,rr,mid+1,y,f,s);

    }

    push_up(i,ll,rr);

}


LL query(int i,int ll,int rr,int x,int y) {
    push_down(i,ll,rr);
    if(ll == x && rr == y) {
        LL ret = 0;
        for(int j = 1; j <= 9; ++j) {
            ret += 1LL*j*sum[i][j];
        }
        return ret;
    }
    if(y <= mid) return query(ls,ll,mid,x,y);
    else if(x > mid) return query(rs,mid+1,rr,x,y);
    else {
        return query(ls,ll,mid,x,mid)+query(rs,mid+1,rr,mid+1,y);
    }
    push_up(i,ll,rr);

}

int n,q;

int main() {

    scanf("%d%d",&n,&q);

    for(int i = 1; i <= n; ++i) {
        scanf("%I64d",&a[i]);
    }

    H[0] = 1;
    for(int i = 1; i <= 13; ++i) H[i] = H[i-1]*10;

    build(1,1,n);

    for(int i = 1; i <= q; ++i) {
        int op,x,y,l,r;
        scanf("%d",&op);
        if(op == 1) {
            scanf("%d%d%d%d",&l,&r,&x,&y);
            if(x == y) continue;
            update(1,1,n,l,r,x,y);
        }
        else {
            scanf("%d%d",&l,&r);
            printf("%I64d\n",query(1,1,n,l,r));
        }
    }

    return 0;
}

Codeforces 794F. Leha and security system 線段樹