/*
求:
......999999999999(n位)
..................
99999999999
9999999999
999999999
99999999
999999
99999
9999
999
99
9
相加
*/
#include <stdio.h>
#include <stdlib.h>
#define N 100
int main()
{
int dight[N*N];
int top=0;
int formerdight=0;
int dightnow=0;
for(int i=1;i<=N;i++){
int dightnow=9*(N+1-i)+formerdight;
dight[top++]=dightnow%10;
formerdight=dightnow/10;
}
while(formerdight!=0){
dight[top++]=formerdight%10;
formerdight=formerdight/10;
}
for(int j=top-1;j>=0;j--)
printf("%d",dight[j]);
return 0;
}