#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
using namespace std;

int d[10000005];

int main() {
int n, x, ans=0;
cin >> n >> x;

d[x]=1;
for (int i=1;i <= n;i++) {
d[i]=d[i%10];
if (i/10) d[i]+=d[i/10];
ans+=d[i];
}

cout << ans;

return 0;
}

dp大法，d[i]=d[个位]+d[各位前面的数];

#include<iostream>
using namespace std;
int main(){
    long long ans=0,x,n,i,temp;
    cin>>n>>x;
    for(i=1;i<=n;++i){
        temp=i;
        do{
            if(temp%10==x) ++ans;
            temp/=10;
        }while(temp);
    }
    cout<<ans;
    return 0;
}

###__DP大法__
```c++
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 39652 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 39648 KiB, score = 10
Accepted, time = 0 ms, mem = 39652 KiB, score = 100
代码
#include <cstdio>
int dp[10000001],n,x,sum = 0;
int main() {
scanf("%d%d",&n,&x);
dp[x] = 1;
for (int i = 1;i <= n;i++) {
dp[i] = dp[i%10];
if (i/10) dp[i] += dp[i/10];
sum += dp[i];
}
printf("%d",sum);
return 0;
}
```

我的pascal代码如下：

var n,i,t:longint;
j,x:integer;
ch:char;
s:string;
begin
t:=0;
read(n,x);
ch:=chr(x+ord('0'));
for i:=1 to n do
begin
str(i,s);
for j:=1 to length(s) do
if ch=s[j] then inc(t);
end;
writeln(t);
end.


#include<iostream>
using namespace std;
int main()
{
int x,s,k,i,t;
cin>>k>>x;
s=0;
for(i=1;i<=k;i++)
{
t=i;
do{
if(t%10==x)
s++;
t=t/10;
}while(t!=0);
}
cout<<s;
return 0;
}

#include <cstdio>
using namespace std;
void count_num(int a,int x)
{
    int count[10] = {0};
    int n = 1;
    int zero = 0;
    int weishu = 0;
    while(a/n)
    {
        weishu++;
        int lowp = a - (a/n)*n;
        int curr = (a/(n))%10;
        int highp = a/(n*10);
        for (int i = 0;i<10;i++)
        {
            if (i<curr)
            {
                count[i] += (highp+1)*n;
            }
            else if (i == curr)
            {
                count[i] += (highp)*n;
                count[i] += (lowp+1);

            }
            else
            {
                count[i] += (highp)*n;
            }
        }
        n *=10;
    }
    int t = 1;
    while(t<=weishu)
    {
        n /=10;
        zero += t*(n-1 -n/10 +1);
        t++;
    }
    count[0] -=zero;
    printf("%d",count[x]);
}
int main(){
    int n,x;
    scanf("%d%d",&n,&x);
    count_num(n,x);
}

感觉像吃了翔，老犯错，最后终于0ms

Pascal超短代码
var n,i,j:longint; s:string; x:char;
a:array ['0'..'9'] of longint;
begin
readln(n,x,x);
for i:=1 to n do
begin
str(i,s);
for j:=1 to length(s) do
inc(a[s[j]]);
end;
write(a[x]);
end.

#include<iostream>
#include<cstdio>
using namespace std;
int main(){
int n,x;
int count=0;
cin>>n>>x;
for(int i=1;i<=n;i++)
{
int k=i;
do{
if(k%10==x)count=count+1;

k=k/10;
}while(k);

}
cout<<count;
return 0;
}

给大家一个数字转字符串的函数
str(x,s);
x---数字 s---字符串
例：str(123,s);
则s='123'

program ym;
var n,m,l,s,i,j:longint;
a,t:string;
begin
readln(n,m);
a:=chr(m+48);
s:=0;
for i:=1 to n do
begin
str(i,t);
l:=length(t);
for j:=1 to l do
if t[j]=a then s:=s+1;
end;
write(s);
end.

var
n,x,i,sum:longint;
s:ansistring;
procedure ok(s:ansistring);
var j:longint;
begin
for j:=1 to length(s) do
if s[j]=chr(x+48) then inc(sum);
end;
begin
readln(n,x);
for i:=1 to n do
begin
str(i,s);
ok(s);
end;
writeln(sum);
end.

过程部分把s[j]=chr(x+48)写成了s[j]='1'，没过，以为longint不行，开int64没去编译过就交上去，结果有没过~~~如此水的题目，我竟然测了三次。。。太失败了

小田君又来发题解啦~~~
简单的计数问题，附代码：
#include <cstdio>

int Count(int n,int x){
int cnt = 0;
while(n != 0){
if(n%10 == x) cnt++;
n /= 10;
}
return cnt;
}

int main(){
int n,x,cnt = 0;
scanf("%d %d",&n,&x);
for(int i = 1; i <= n; i++) cnt += Count(i,x);
printf("%d",cnt);

return 0;
}

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int cnt=0;
int getx(int x,int i){
int cnt1=0;
while(i>0){
if(i%10==x)
cnt1++;
i/=10;
}
return cnt1;
}
int main(){
int n,x;
cin>>n>>x;
for(int i=1;i<=n;i++)
cnt+=getx(x,i);
cout<<cnt;
return 0;
}

var n,x,i,j,num:longint; s:string;
begin
readln(n,x); num:=0;
for i:=1 to n do
begin
str(i,s);
for j:=1 to length(s) do
if s[j]=chr(48+x) then
inc(num);
end;
writeln(num);
end.

【水】【C】
#include<stdio.h>
int main()
{
int n,x,i,j,a,ans=0,mod=1;
scanf("%d%d",&n,&x);
for(i=1;i<=n;i++)
{

a=i;
while(a>0)
{
if(a%10==x) ans++;
a=a/10;
}
}
printf("%d",ans);
return 0;
}

正则表达式就是好用

---

import java.io.*;
import java.util.*;
import java.util.regex.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final int end=sc.nextInt();
        int count=0;
        Pattern p=Pattern.compile(sc.next());
        for (int i = 1; i <=end; i++) {
            Matcher m=p.matcher(Integer.toString(i));
            while(m.find())
                count++;
        }
        System.out.println(count);
    }
}

整个题目归功于sprintf
c
#include <stdio.h>
int main(){
int n,num,i,j,ans = 0;
scanf("%d %d",&n,&num);
char s[1000010];
for(i = 1;i<=n;i++){
sprintf(s+1, "%d",i);
for(j = 1;s[j];j++)
if(s[j] == num+48) ans++;
}
printf("%d\n",ans);
return 0;
}


题解

```
/* ***********************************************
Author :guanjun
Created Time :2016/2/18 11:03:23
File Name :vijosp1848.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
return a>b;
}
int main()
{
#ifndef ONLINE_JUDGE
//freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int n,x,a;
while(cin>>n>>x){
ll num=0;
for(int i=1;i<=n;i++){
a=i;
//cout<<a<<endl;
while(a>0){
if(a%10==x)num++;
a/=10;
}
}
printf("%lld\n",num);
}
return 0;
}

program p1848;
var s:string;
题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题
题题题题题题题题题题题题题题题题题题题题题题题题题题题题水水水水题题题题题
题题题题题题题题题题题题题题题题题题题题题题题题水水水水水水水水水题题题题
题题题题题题题题题题题题题题题题题题水水水水水水水水水水水水水水水题题题题
题题题题题题题题题题题题题题题题水水水水水水水水水水水水水水水题题题题题题
题题题题题题题题题题题水水水水题水水水水水水水水水水水题题题题题题题题题题
题题题题题题题题水水水水水水水题水水水题题水水水水水题题题题题题题题题题题
题题题水水水水水水水水水水水水题题题题题题水水水水题题题题题题题题题题题题
题水水水水水水水水水水水水水水题题题题题题水水水水题题题题题题题题题题题题
题水水水水水水水水水水水水题题题题题题题水水水水水水水水水水水题题题题题题
题水水水水水水水水水水水水题题题题题题水水水水水水水水水水水水水水题题题题
题题水水水水水水水水水水题题题题题水水水水水水题题题水水水水水水水题题题题
题题题题题题题题水水水水题题题题题水水水水题题题题题题水水水水水题题题题题
题题题题题题题题水水水水题题题题水水水水题题水水题题题水水水水水题题题题题
题题题题题题题题水水水水题题题题水水水水题题水水水水题水水水水水题题题题题
题题题题题题题题水水水水题题题题水水水水题题水水水水题水水水水水题题题题题
题题题题题题题题水水水水题题题题水水水水题题水水水题题水水水水水题题题题题
题题题题题题题题水水水水题题题题水水水水题题水水水题题水水水水水题题题题题
题题题题题题题题水水水水题题题题水水水水题水水水水题题水水水水水题题题题题
题题题题题题题题水水水水题题题题水水水水题水水水水题题水水水水水题题题题题
题题题题题题题题水水水水题题题题水水水水题水水水水题题水水水水水题题题题题
题题题题题题题题水水水水题题题题水水水水题水水水水题题水水水水水题题题题题
题题题题题题题题水水水水题题题题水水水题题水水水水题题水水水水水题题题题题
题题水水题题题水水水水水题题题题水水水题题水水水题题题水水水水水题题题题题
题题水水水水水水水水水水题题题题题水水题题水水题题题题水水水水水题题题题题
题题题水水水水水水水水水题题题题题题题题水水水题题题题水水水水题题题题题题
题题题题题水水水水水水水题题题题题题题题水水水题水水水水题题题题题题题题题
题题题题题题水水水水水水题题题题题题题水水水水题题水水水水题题题题题题题题
题题题题题题题题题水水水题题题题题题水水水水水题题题水水水水水水水题题题题
题题题题题题题题题题题题题题题题水水水水水水题题题题题水水水水水水题题题题
题题题题题题题题题题题题题题题水水水水水水题题题题题题水水水水水水水题题题
题题题题题题题题题题题题题题水水水水水题题题题题题题题题水水水水水水题题题
题题题题题题题题题题题题题水水水水水题题题题题题题题题题题水水水水题题题题
题题题题题题题题题题题题水水水题题题题题题题题题题题题题题题水水水题题题题
题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题题
c:char;

n:longint;
t: int64;
a,i,j:longint;
begin
readln(n,a);
str(a,s);
c:=s[1];t:=0;

for i:=1 to n do

begin

str(i,s);

for j:=1 to length(s) do
if s[j]=c then t:=t+1;
end;
writeln(t);
end.

编译成功

Free Pascal Compiler version 2.6.2 [2013/02/12] for i386
Copyright (c) 1993-2012 by Florian Klaempfl and others
Target OS: Win32 for i386
Compiling foo.pas
Linking foo.exe
14 lines compiled, 0.1 sec , 28032 bytes code, 1628 bytes data

测试数据 #0: Accepted, time = 15 ms, mem = 4528 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 4528 KiB, score = 10

测试数据 #2: Accepted, time = 46 ms, mem = 4528 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 4528 KiB, score = 10

测试数据 #4: Accepted, time = 62 ms, mem = 4528 KiB, score = 10

测试数据 #5: Accepted, time = 31 ms, mem = 4528 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 4528 KiB, score = 10

测试数据 #7: Accepted, time = 46 ms, mem = 4528 KiB, score = 10

测试数据 #8: Accepted, time = 78 ms, mem = 4528 KiB, score = 10

测试数据 #9: Accepted, time = 109 ms, mem = 4528 KiB, score = 10

Accepted, time = 387 ms, mem = 4528 KiB, score = 100

代码
var j,i,n,x:longint;
a:array[1..1000000]of longint;
begin
readln(n,x);
j:=0;
for i:=1 to n do a[i]:=i;
for i:= 1 to n do if a[i]=x then j:=j+1
else while a[i]<>0 do
begin
if a[i] mod 10=x then j:=j+1;
a[i]:=a[i] div 10;
end;
writeln(j);
end.