Description

         有一位售票员给乘客售票,对于每位乘客,他会卖出多张连续的票,直到已卖出的编号的所有位置上的数的和不小于给定的正数k。然后他会按照相同的规则给下一位乘客售票。初始时,售票员持有的编号是从L到R的连续整数。请你求出,售票员可以售票给多少位乘客。

Input

         三个整数L,R,k。

Output

         一个正整数,表示能够拿到票的乘客数。
f[a][b][c]表示开头的和为b,末尾在[0,10a)的数,前面补上和为c的一段数字,能分出的段数和剩余的和
预处理a=0..19,b=0..190,c=0..k-1的答案
然后类似zkw线段树查出区间[l,r]对应的信息,合并答案
#include<cstdio>
typedef long long i64;
i64 l,r,p10[];
struct state{
i64 c;
int r;
void operator+=(state w){
c+=w.c;
r=w.r;
}
}f[][][];
struct pos{
int a,b;
}stk1[],stk2[];
int stp1=,stp2=;
int p1=,p2=;
int k,sl[],pl=,sr[],pr=;
int main(){
scanf("%lld%lld%d",&l,&r,&k);
p10[]=;
for(int i=;i<=;++i)p10[i]=p10[i-]*;
for(int j=;j<=;++j){
for(int a=;a<k;++a){
f[][j][a]=(state){a+j>=k,a+j>=k?:a+j};
}
}
for(int i=;i<=;++i){
for(int j=;j<=-i*;++j){
for(int a=;a<k;++a){
state&w=f[i][j][a]=f[i-][j][a];
for(int b=;b<;++b){
w+=f[i-][j+b][w.r];
}
}
}
}
--l;++r;
while(l)sl[++pl]=l%,l/=;
while(r)sr[++pr]=r%,r/=;
pl=pr;
int eq=pr;
while(sl[eq]==sr[eq])--eq;
int cl=,cr=;
for(int i=;i<=pr;++i)cl+=sl[i],cr+=sr[i];
for(int i=;i<eq;++i){
cl-=sl[i],cr-=sr[i];
for(int a=sl[i]+;a<=;++a)stk1[stp1++]=(pos){i-,cl+a};
for(int a=sr[i]-;a>=;--a)stk2[stp2++]=(pos){i-,cr+a};
}
cr-=sr[eq];
for(int a=sl[eq]+;a<sr[eq];++a)stk1[stp1++]=(pos){eq-,cr+a};
while(stp2)stk1[stp1++]=stk2[--stp2];
state w=(state){,};
for(int i=;i<stp1;++i)w+=f[stk1[i].a][stk1[i].b][w.r];
printf("%lld\n",w.c);
return ;
}

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