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

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
    int maxNum;
    int correntNum;
    int counter = 0;
    int findNum;
    int num;
    int temp;
    scanf("%d%d",&maxNum,&findNum);
    for(correntNum = 1; correntNum <= maxNum; correntNum++)
    {
        temp = correntNum;
        do
        {
            num = temp % 10;
            temp = temp / 10;
            if(num == findNum)
                counter++;
        }while(temp > 0);
    }
    printf("%d",counter);
    return 0;
}

国服影流之主来了！！！
var
n,t,l,k,num,i:longint;
s,num1:string;
begin
read(n,num);
str(num,num1);
for i:=1 to n do
begin
str(i,s);
k:=length(s);
l:=0;
while l<k do
begin
inc(l);
if s[l]=num1 then inc(t);
end;
end;
write(t);
end.

#include<bits/stdc++.h>
using namespace std;
int main(){
int n,x,sum=0,m;
cin>>n >>x;
for(int i=1;i<=n;i++){
int y=i;
do{
m=y%10;
if(m==x)
sum++;
y=y/10;
}while(y>=1);
}
cout<<sum;
return 0;
}

不用对\(1,\cdots,n\)的暴力数位遍历。按\(10^m\)将\(1,\cdots,n\)划分为若干个组，每组对应个位数、十位数等数位中出现的个数，通过数学计算直接加总。复杂度从\(O(n\log(n))\)降到\(O(\log(n))\)。

被数学函数坑了，忘记了有\(log10\)，用换底公式结果出现了
cpp
floor(log(1000)/log(10))=2
的情况……哪位路过大神还请能解释下这是为什么，万谢！

以下是代码：

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int n, x, did, m;
    cin >> n >> x;
    did = 0; // did is appearance times of x
    m = (int)floor(log10(n)); // m is the digit capacity of n
                              // Note: floor(log(n) / log(10)) will go wrong when n=1e3 because floor causes loss in division!!
                              //cout << log(n) / log(10) << " " << floor(log(n) / log(10)); // results: 3 2
    if (x) // 0 is a little bit tricky
    {
        for (int i = 1; i <= m; i++) // for complete 10^i cycle
            did += n / (int)pow(10, i) * (int)pow(10, i - 1);
        did += (n % 10 >= x) ? 1 : 0;   // first digit
        for (int i = 1; i <= m; i++)    // other digits
            if ((n / (int)pow(10, i)) % 10 > x)
                did += (int)pow(10, i);
            else if ((n / (int)pow(10, i)) % 10 == x)
                did += n % (int)pow(10, i) + 1;
    }
    else
    {
        for (int i = 1; i <= m; i++) // for complete 10^i cycle
            did += (n / (int)pow(10, i) - 1) * (int)pow(10, i - 1) + 1;
        // no first digit's contribution
        for (int i = 1; i < m; i++) // other digits
            if ((n / (int)pow(10, i)) % 10 > 0) // not 0 at i+1-th position
                did += (int)pow(10, i) - 1;
            else if ((n / (int)pow(10, i)) % 10 == 0) // there is a 0 at i+1-th position
                did += n % (int)pow(10, i); // not +1 as the non-zero number case
    }

    cout << did;

    return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main(){
int n,m,t,ts;

cin>>n>>m;
for(int i=1;i<=n;i++){
ts=i;
while(ts){
if(ts%10==m)
t++;
ts=ts/10;

}
}
cout<<t;
return 0;
}

数据很水，只要一个模拟就可以通过了

AC code：

#include<bits/stdc++.h>
using namespace std;
int main(){
    int a,m,t=0,d;
    cin>>a>>m;
    for(int i=1;i<=a;i++)
    {
        d=i;
        while(d>=10){
            if(d%10==m) t++;
            d=d/10; 
        }
        if(d==m) t++;
    }
    cout<<t;
    return 0;
 } 

简单
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,x,sz[1000001];
long long sum=0;
cin>>n>>x;
for(int i=1;i<=n;i++)
{
sz[i]=i;
}
for(int i=1;i<=n;i++)
{
while(sz[i]!=0)
{
int g=sz[i]%10;
sz[i]=sz[i]/10;
if (g==x) sum++;
}
}
cout<<sum<<endl;
return 0;
}

#include <iostream>
using namespace std;
/*
试计算在区间 1 到 n 的所有整数中,数字 x(0 ≤ x ≤ 9)共出现了多少次?
例如,在 1 到 11 中,即在 1、2、3、4、5、6、7、8、9、10、11 中,数字 1 出现了 4 次。
*/
int main()
{
int i, n, x, k = 0, j;
scanf_s("%d %d", &n, &x);
for (i = 1; i <= n; i++) {
j = i;
while (j % 10 == x) {
k++;
j = j / 10;
}
}
printf("%d", k);
system("pause");
return 0;
}

#1 Accepted 3ms 264.0 KiB
#2 Accepted 2ms 256.0 KiB
#3 Accepted 556ms 380.0 KiB
#4 Accepted 23ms 256.0 KiB
#5 Accepted 690ms 384.0 KiB
#6 Accepted 412ms 348.0 KiB
#7 Accepted 3ms 340.0 KiB
#8 Accepted 367ms 380.0 KiB
#9 Accepted 679ms 256.0 KiB
#10 Accepted 664ms 364.0 KiB

#include <iostream>
#include <algorithm>
#include <cstring>
#include <sstream>
using namespace std;
int main()
{
string s;
int n,sum=0,x;
cin>>n>>x;
for(int i = 1;i <= n;i++)
{
stringstream ss;
ss<<i;
ss>>s;
for(int j = 0;j <= s.length();j++)
if(s[j] == x+0x30) sum++;
}
cout<<sum;
return 0;
}

<?php
/**
 * Created by PhpStorm.
 * User: Administrator
 * Date: 2018\5\14 0014
 * Time: 18:09
 */

//试计算在区间 1 到 n 的所有整数中,数字 x(0 ≤ x ≤ 9)共出现了多少次?例如,在 1 到 11 中,即在 1、2、3、4、5、6、7、8、9、10、11 中,数字 1 出现了 4 次。

class count{
    protected $min = 1;
    protected $max;
    protected $num;

    public function outPutCount(){
        $count = 0;
        for ($i = $this->min; $i <= $this->max; $i ++){
            $count = $count + substr_count($i,$this->num);
        }
        return $count;
    }

    public function setMax($max){
        $this->max = $max;
        return $this;
    }

    public function setNum($num){
        $this->num = $num;
        return $this;
    }

}

$stdin = fopen('php://stdin','r');
$data = trim(fgets($stdin));
$lists = explode(' ',$data);
$max = $lists[0];
$num = $lists[1];
$counter = new count();
$count = $counter->setMax($max)->setNum($num)->outPutCount();
echo $count;

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int ans=0,n,x;
    scanf("%d%d",&n,&x);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=10000000;j=j*10)
            if(i>=j)if(i/j%10==x)ans++;
    printf("%d",ans);
    return 0;
}

参考刘汝佳《算法竞赛入门经典》。
用sprintf
#include<stdio.h>
#include<string.h>
char s[15];
int main()
{
int n,x;
scanf("%d%d",&n,&x);
int ans=0;
for(int i=1;i<=n;i++)
{
sprintf(s,"%d",i);
for(int j=0;j<strlen(s);j++)
if(s[j]-'0'==x)ans++;
}
printf("%d\n",ans);
return 0;
}