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 numbers a1, a2, ..., an describes the amount of money each client has. Leha wants to make requests to the database of the bank, finding out the total amount of money on some subsegments of the sequence and changing values of the sequence on some subsegments. Using a bug in the system, Leha can requests two types of queries to the database:
- 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.
题意：
给你n个数
操作1：l r x y，区间[l,r]内所有数，数位上为x的都转化为y
操作2: l r 求区间和
题解：
线段树区间合并
建立10颗线段树，分别表示数字0~9所代表的值
将x转化为y也就是在将第x颗线段树区间[l,r]和减去，加到第y颗线段树上
这里的延时操作有点小技巧
每次push_down的时候保持每个点(0~9)指向唯一的另外一个点，这样再更新的时候才不会超时
#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 ",query(1,1,n,l,r)); } } return 0; }