#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;
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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.

var
a,b,c,x,y,z,i,l,j:longint;
begin
readln(a,b);
for i:=1 to a do
begin
z:=i;
repeat
x:=z mod 10 ;
y:=z div 10 ;
if x=b then inc(l);
z:=y;
until y=0;
end;
writeln(l);
end.