var
i,j,k,m,n,p,ans:longint;
x:string;
a:array[1..1000000]of string;
begin
read(n,m);
str(m,x);
for i:=1 to n do str(i,a[i]);
for i:=1 to n do
begin
k:=length(a[i]);
for j:=1 to k do
begin
p:=pos(x,a[i]);
if p<>0 then inc(ans);
delete(a[i],1,p);
end;
end;
writeln(ans);
end.

use std::io;

pub fn main() {
let mut input1 = String::new();
let n :i32;
let x :i32;
let mut k :i32 = 0;
let mut j :i32;

match io::stdin().read_line(&mut input1){
Ok(n) => {
//println!("{} bytes read",n);
//println!("{}",input1);
}
Err(error) => println!("error:{}", error),
}

let v:Vec<&str> = input1.trim().split(' ').collect();
n = v[0].parse::<i32>().unwrap();
x = v[1].parse::<i32>().unwrap();
for i in 1..(n+1) {
j=i;
while j != 0 {
if j%10 == x { k+=1;}
j = j/10;
}
}
println!("{}",k);
//println!("input:{}",x);
}

use std::io;

pub fn main() {
let mut input1 = String::new();
let n :i32;
let x :i32;
let mut k :i32 = 0;
let mut j :i32;

match io::stdin().read_line(&mut input1){
Ok(n) => {
//println!("{} bytes read",n);
//println!("{}",input1);
}
Err(error) => println!("error:{}", error),
}

let v:Vec<&str> = input1.trim().split(' ').collect();
n = v[0].parse::<i32>().unwrap();
x = v[1].parse::<i32>().unwrap();
for i in 1..(n+1) {
j=i;
while j != 0 {
if j%10 == x { k+=1;}
j = j/10;
}
}
println!("{}",k);
//println!("input:{}",x);
}

蛮简单的。。。一遍AC。。。自己参考
#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
int n,x,r,q,num=0;
scanf("%d%d",&n,&x);
for(int i=1;i<=n;i++)
{
int temp=i;
do
{
r=temp%10;
q=temp/10;
if(r==x) num++;
temp=q;
}while(q!=0);
}
cout<<num;
}

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

#include<cstdio>
int main()
{
int num=0;
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
int j=i;
while(j>=1)
j%10==m?num++,j/=10:j/=10;
}
printf("%d",num);
}

#include<cstdio>
int main()
{
int n,m=0;
scanf("％d",&n);
while(n!="#")
{
if(n/10=1)
m+=1;
scanf("%d",&n);
}
printf("%d",m);
return 0;
}

pascal 神奇的字符串做法

var n,x,l,i,j,t:longint;s1,s:string;
begin
    read(n,x);str(x,s1);t:=0;
    for i:= 1 to n do begin
        str(i,s);l:=length(s);
        j:=0;
        repeat
            j:=j+1;
            if s[j]=s1 then t:=t+1;
        until j=l;
    end;
    write(t);
end.

我的pascal
var
n,x,i,y,z:longint;
begin
y:=0;
read(n,x);
for i:= 1 to n do
begin
z:=i;
while z>0 do
begin
if z mod 10=x
then y:=y+1;
z:=z div 10;
end;
end;
write(y);
end.

#include<stdio.h>
int main()
{
    int num=0;
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        int j=i;
        while(j>=1)
        j%10==m?num++,j/=10:j/=10;
    }
    printf("%d",num);
}

#include<stdio.h>
int main()
{
    int num=0;
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        int j=i;
        while(j>=1)
        j%10==m?num++,j/=10:j/=10;
    }
    printf("%d",num);
}

Python AC
Python
a,b = raw_input().split()
count = 0
for i in range(1,int(a)+1):
count += str(i).count(b)
print count


//
// Created by JimmyChen on 6/23/17.
//

#include <cstdio>
#include <iostream>

using namespace std;
int a, num;

int main(){
    cin >> a >> num;
    int count = 0;
    for (int i = 1; i <= a; i++){
        int temp = i;
        while (temp > 0){
            if (temp % 10 == num){count++;}
            temp = temp / 10;
        }
    }
    printf("%d", count);
}

#include <iostream>
#include <cmath>

using namespace std;

int main() {
int n = 0, x = 0;
cin >> n >> x;
int dig = 1, nowdig = 0, res = 0, lev = 0, num = n, zlev = 0;
while (n>0) {
nowdig = n % 10;
if (!lev) {
if (nowdig >= x&&n>10) {
res = 1;
}
else res = 0;
}
else {
if (n < 10 && !x) {
res += (nowdig - 1)*lev+zlev;
}
else {
res += nowdig*lev;
if (nowdig > x)res += pow(10, dig - 1);
else if (nowdig == x) {
int tmp = pow(10, dig);
if (x > 0) {
res += (num%tmp - nowdig*tmp / 10 + 1);
}
else {
res += num%tmp+1;
}
}
}
}
if (!x)zlev += 9 * lev;
lev *= 10;
lev += pow(10, dig - 1);
dig++;
n /= 10;
}
cout << res << "\n";
return 0;
}

#include <cstdio>
#include <string>
#include <sstream>

using namespace std;
/*
    Created and edited by Ray Eldath.
*/
int main() {
    int r[10];
    for (int i=0; i<10; i++)
        r[i]=0;
    int a, b;
    scanf("%d%d", &a, &b);
    stringstream ss;
    for (int i = 1; i <= a; i++)
        ss << i;
    string s = ss.str();
    for (int i = 0; i < s.size(); i++)
        r[s[i] - '0']++;
    printf("%d", r[b]);
}

这题似乎没什么好说的，大概就是用模运算加判定吧。
主要是这题的题目的检测有问题，不然我这代码应该过不了，比如n给一个1000，我这代码就炸了……

#include <iostream>

using namespace std;

int main()
{
    int i,n,x,k=0,j; //k为总数 
    cin >>n>>x;
    for(i=1;i<=n;i++)
    {
        j=i;
        while(j)
        {
            if(j%10==x)
            k++;
            j=j/10;
        }
    }
    cout<<k;
    return 0;
}

这是一道水题，只要跟着题目要求模拟一遍就OK了，好了不多说了，直接上代码
#include <bits/stdc++.h>
using namespace std;
int n, m, a;
int cnt = 0;
int main()
{
cin >> n >> m;
for(int i = 1; i <= n; ++i)
{
a = i;
while(a > 0)
{
if(a % 10 == m)
{
cnt++;
}
a = a / 10;
continue;
}
}
cout << cnt << endl;
return 0;
}

#include <stdio.h>
int main()
{
int i,n,x,k=0,j;
scanf("%d %d",&n,&x);
for(i=1;i<=n;i++)
{
j=i;
while(j)
{
if(j%10==x)
k++;
j=j/10;
}
}
printf("%d",k);
return 0;
}

#include <cstdio>
#include <iostream>
using namespace std;
int main() {
int i,j,x,n;
int a[10]={0};
cin>>n>>x;
for(i=1;i<=n;i++){
j=i;
while(j>0){
a[j%10]++;
j/=10;
}

}
cout<<a[x];
return 0;
}

用c++写的，思路是用一个递归函数处理每个数，然后用for循环处理全部的数

#include <iostream>

long int count(long int temp, long int x);

int main()
{
    long int n, x, result=0;
    std::cin >> n >> x;
    for(long int i=1;i<=n;i++)
    {
        result+=count(i, x);
    }

    std::cout << result << std::endl;

    return 0;
}

long int count(long int temp, long int x)
{
    if(temp/10 < 1)
    {
        if(temp == x)
        {
            return 1;
        }else
        {
            return 0;
        }
    }else
    {
        if(temp%10 == x)
        {
            return 1 + count(temp/10, x);
        }else
        {
            return 0 + count(temp/10, x);
        }
    }
}