Problem Statement

E - Replace Digits

You have a string SS of length NN. Initially, all characters in SS are 1s.

You will perform queries QQ times. In the ii-th query, you are given two integers L_i, R_iLi​,Ri​ and a character D_iDi​ (which is a digit). Then, you must replace all characters from the L_iLi​-th to the R_iRi​-th (inclusive) with D_iDi​.

After each query, read the string SS as a decimal integer, and print its value modulo 998,244,353998,244,353.

Constraints

• 1 \leq N, Q \leq 200,0001≤N,Q≤200,000
• 1 \leq L_i \leq R_i \leq N1≤Li​≤Ri​≤N
• 1 \leq D_i \leq 91≤Di​≤9
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

NN QQ
L_1L1​ R_1R1​ D_1D1​
::
L_QLQ​ R_QRQ​ D_QDQ​

Output

Print QQ lines. In the ii-th line print the value of SS after the ii-th query, modulo 998,244,353998,244,353.

Sample Input 1 Copy

Copy

8 5
3 6 2
1 4 7
3 8 3
2 2 2
4 5 1

Sample Output 1 Copy

Copy

11222211
77772211
77333333
72333333
72311333

题意：有一个长为n的串，初始的每个位置都是1

m个操作，每次操作l r x，把l~r这一段位置上的每个数都改成x

每次修改之后看总的串膜上998,244,353之后的结果

思路：先建树

sum表示l~r这段的总数，s表示l~r这段的数应该乘的数位，lazy表示懒标记

建树的时候sum和s随着更新

但是修改的时候只有sum在更新，因为l~r这段应该乘的位数是不变的

#include
using namespace std;
const int N=2e5+10;
const long long mod=998244353;
typedef long long ll;
struct name{int l,r;ll sum,lazy,s;
}tr[N*4];
int n,m;
ll a[N];
void pushup(int u){tr[u].sum =(tr[u<<1].sum+tr[u<<1|1].sum  )%mod;tr[u].s =(tr[u<<1].s +tr[u<<1|1].s)%mod;
}
void build(int u,int l,int r){if(l==r){tr[u]={l,r,a[l],0,a[l]};}else{int mid=l+r>>1;tr[u]={l,r,0,0,0};build(u<<1,l,mid);build(u<<1|1,mid+1,r);pushup(u);}
}
void pushdown(int u){if(tr[u].lazy ){tr[u<<1].lazy =tr[u].lazy ;tr[u<<1].sum =tr[u<<1].s *tr[u].lazy %mod;tr[u<<1|1].lazy =tr[u].lazy ;tr[u<<1|1].sum =tr[u<<1|1].s *tr[u].lazy %mod;tr[u].lazy =0;		}
}
void modify(int u,int l,int r,ll x){if(l<=tr[u].l &&r>=tr[u].r ){tr[u].lazy =x;tr[u].sum =tr[u].s *x%mod;}else{pushdown(u);int mid=tr[u].l +tr[u].r >>1;if(l<=mid)modify(u<<1,l,r,x);if(r>mid)modify(u<<1|1,l,r,x);tr[u].sum =(tr[u<<1].sum +tr[u<<1|1].sum)%mod; }
}
int main(){scanf("%d%d",&n,&m);a[n]=1;for(int i=n-1;i>=1;i--){a[i]=a[i+1]*10%mod;}build(1,1,n);while(m--){int l,r;ll x;scanf("%d%d%lld",&l,&r,&x);modify(1,l,r,x);printf("%lld\n",tr[1].sum );}return 0;
}