Buggy python submission (verdict: Time Limit Exceeded):
L,A,B,mod=map(int,input().split())

def keta(k):# k-桁の数の初項、末項、項数を求める
    if A>=10**k:
        return 0,0,0
    begin=10**(k-1)
    end=10**k

    if begin>A+B*(L-1):
        return 0,0,0

    sh=A+max(-1,(begin-1-A)//B)*B+B
    ma=min(A+max(0,(end-1-A)//B)*B,A+B*(L-1))
    kou=(ma-sh)//B+1
    
    return sh,ma,kou

def f(n,r): # 1+r+...+r**(n-1)
    if n==1:
        return 1
    if n%2==0:
        return (f(n//2,r) + pow(r,n//2,mod)*f(n//2,r))%mod
    return (r*f(n-1,r)+1)%mod

def g(n,r): # 0+r+2*r**2+...+(n-1)*r**(n-1)
    if n==1:
        return 0
    if n%2==0:
        return g(n//2,r)+pow(r,n//2,mod)*(g(n//2,r)+n//2*f(n//2,r))
    return r*g(n-1,r)+r*f(n-1,r)

keta_count=1
ANS=0
for i in range(18,0,-1):
    sh,ma,kou = keta(i)
    
    if kou<=0:
        continue
    

    #print(i,sh,ma,kou)

    ANS=(keta_count*(ma*f(kou,10**i)-B*g(kou,10**i))+ANS)%mod
    #print(ANS)
    keta_count = keta_count * pow(10,kou * i,mod)

print(ANS)
    
    


Fixed submission:
L,A,B,mod=map(int,input().split())

def keta(k):# k-桁の数の初項、末項、項数を求める
    if A>=10**k:
        return 0,0,0
    begin=10**(k-1)
    end=10**k

    if begin>A+B*(L-1):
        return 0,0,0

    sh=A+max(-1,(begin-1-A)//B)*B+B
    ma=min(A+max(0,(end-1-A)//B)*B,A+B*(L-1))
    kou=(ma-sh)//B+1
    
    return sh,ma,kou

from functools import lru_cache

@lru_cache(maxsize=None)
def f(n,r): # 1+r+...+r**(n-1)
    if n==1:
        return 1
    if n%2==0:
        return (f(n//2,r) + pow(r,n//2,mod)*f(n//2,r))%mod
    return (r*f(n-1,r)+1)%mod

@lru_cache(maxsize=None)
def g(n,r): # 0+r+2*r**2+...+(n-1)*r**(n-1)
    if n==1:
        return 0
    if n%2==0:
        return g(n//2,r)+pow(r,n//2,mod)*(g(n//2,r)+n//2*f(n//2,r))
    return r*g(n-1,r)+r*f(n-1,r)

keta_count=1
ANS=0
for i in range(18,0,-1):
    sh,ma,kou = keta(i)
    
    if kou<=0:
        continue
    

    #print(i,sh,ma,kou)

    ANS=(keta_count*(ma*f(kou,10**i)-B*g(kou,10**i))+ANS)%mod
    #print(ANS)
    keta_count = keta_count * pow(10,kou * i,mod)

print(ANS)

Buggy python submission (verdict: Time Limit Exceeded):
L,A,B,mod=map(int,input().split())

def keta(k):# k-桁の数の初項、末項、項数を求める
    if A>=10**k:
        return 0,0,0
    begin=10**(k-1)
    end=10**k

    if begin>A+B*(L-1):
        return 0,0,0

    sh=A+max(-1,(begin-1-A)//B)*B+B
    ma=min(A+max(0,(end-1-A)//B)*B,A+B*(L-1))
    kou=(ma-sh)//B+1
    
    return sh,ma,kou

def f(n,r): # 1+r+...+r**(n-1)
    if n==1:
        return 1
    if n%2==0:
        return (f(n//2,r) + pow(r,n//2,mod)*f(n//2,r))%mod
    return (r*f(n-1,r)+1)%mod

def g(n,r): # 0+r+2*r**2+...+(n-1)*r**(n-1)
    if n==1:
        return 0
    if n%2==0:
        return g(n//2,r)+pow(r,n//2,mod)*(g(n//2,r)+n//2*f(n//2,r))
    return r*g(n-1,r)+r*f(n-1,r)

keta_count=0
ANS=0
for i in range(18,0,-1):
    sh,ma,kou = keta(i)
    
    if kou<=0:
        continue
    

    #print(i,sh,ma,kou)

    ANS=(pow(10,keta_count,mod)*(ma*f(kou,10**i)-B*g(kou,10**i))+ANS)%mod
    #print(ANS)
    keta_count += kou * i

print(ANS)
    
    


Fixed submission:
L,A,B,mod=map(int,input().split())

def keta(k):# k-桁の数の初項、末項、項数を求める
    if A>=10**k:
        return 0,0,0
    begin=10**(k-1)
    end=10**k

    if begin>A+B*(L-1):
        return 0,0,0

    sh=A+max(-1,(begin-1-A)//B)*B+B
    ma=min(A+max(0,(end-1-A)//B)*B,A+B*(L-1))
    kou=(ma-sh)//B+1
    
    return sh,ma,kou

from functools import lru_cache

@lru_cache(maxsize=None)
def f(n,r): # 1+r+...+r**(n-1)
    if n==1:
        return 1
    if n%2==0:
        return (f(n//2,r) + pow(r,n//2,mod)*f(n//2,r))%mod
    return (r*f(n-1,r)+1)%mod

@lru_cache(maxsize=None)
def g(n,r): # 0+r+2*r**2+...+(n-1)*r**(n-1)
    if n==1:
        return 0
    if n%2==0:
        return g(n//2,r)+pow(r,n//2,mod)*(g(n//2,r)+n//2*f(n//2,r))
    return r*g(n-1,r)+r*f(n-1,r)

keta_count=1
ANS=0
for i in range(18,0,-1):
    sh,ma,kou = keta(i)
    
    if kou<=0:
        continue
    

    #print(i,sh,ma,kou)

    ANS=(keta_count*(ma*f(kou,10**i)-B*g(kou,10**i))+ANS)%mod
    #print(ANS)
    keta_count = keta_count * pow(10,kou * i,mod)

print(ANS)

Buggy python submission (verdict: Time Limit Exceeded):
L,A,B,mod=map(int,input().split())

def keta(k):# k-桁の数の初項、末項、項数を求める
    if A>=10**k:
        return 0,0,0
    begin=10**(k-1)
    end=10**k

    if begin>A+B*(L-1):
        return 0,0,0

    sh=A+max(-1,(begin-1-A)//B)*B+B
    ma=min(A+max(0,(end-1-A)//B)*B,A+B*(L-1))
    kou=(ma-sh)//B+1
    
    return sh,ma,kou

def f(n,r): # 1+r+...+r**(n-1)
    if n==1:
        return 1
    if n%2==0:
        return (f(n//2,r) + pow(r,n//2,mod)*f(n//2,r))%mod
    return (r*f(n-1,r)+1)%mod

def g(n,r): # 0+r+2*r**2+...+(n-1)*r**(n-1)
    if n==1:
        return 0
    if n%2==0:
        return g(n//2,r)+pow(r,n//2,mod)*(g(n//2,r)+n//2*f(n//2,r))
    return r*g(n-1,r)+r*f(n-1,r)

keta_count=0
ANS=0
for i in range(18,0,-1):
    sh,ma,kou = keta(i)
    
    if kou<=0:
        continue
    

    #print(i,sh,ma,kou)

    ANS=(pow(10,keta_count,mod)*(ma*f(kou,10**i)-B*g(kou,10**i))+ANS)%mod
    #print(ANS)
    keta_count += kou * i

print(ANS)
    
    


Fixed submission:
L,A,B,mod=map(int,input().split())

def keta(k):# k-桁の数の初項、末項、項数を求める
    if A>=10**k:
        return 0,0,0
    begin=10**(k-1)
    end=10**k

    if begin>A+B*(L-1):
        return 0,0,0

    sh=A+max(-1,(begin-1-A)//B)*B+B
    ma=min(A+max(0,(end-1-A)//B)*B,A+B*(L-1))
    kou=(ma-sh)//B+1
    
    return sh,ma,kou

from functools import lru_cache

@lru_cache(maxsize=None)
def f(n,r): # 1+r+...+r**(n-1)
    if n==1:
        return 1
    if n%2==0:
        return (f(n//2,r) + pow(r,n//2,mod)*f(n//2,r))%mod
    return (r*f(n-1,r)+1)%mod

@lru_cache(maxsize=None)
def g(n,r): # 0+r+2*r**2+...+(n-1)*r**(n-1)
    if n==1:
        return 0
    if n%2==0:
        return g(n//2,r)+pow(r,n//2,mod)*(g(n//2,r)+n//2*f(n//2,r))
    return r*g(n-1,r)+r*f(n-1,r)

keta_count=1
ANS=0
for i in range(18,0,-1):
    sh,ma,kou = keta(i)
    
    if kou<=0:
        continue
    

    #print(i,sh,ma,kou)

    ANS=(keta_count*(ma*f(kou,10**i)-B*g(kou,10**i))+ANS)%mod
    #print(ANS)
    keta_count = keta_count * pow(10,kou * i,mod)

print(ANS)