#include<iostream>
#include<cstdio>
using namespace std;

int main() {
int n, num, i, t, a;
int sum = 0;
scanf("%d %d", &n, &num);
t = n;
for(i = 1; i <= n; i++) {
a = i;
while(a) {
if(a%10==num)
sum++;
a /= 10;
}
}
printf("%d", sum);
}

纯模拟题，建议C、C++蒟蒻练手。

#include<bits/stdc++.h>
using namespace std ;
//Vijos P1848

int x , y , ans ;
int num(int n)
{
    int ans = 0 ;
    while (n > 0)//枚举n的每一位上的数
    {
        if (y == n % 10)
            ans ++ ;//求个数
        n /= 10 ;
    }
    return ans ;//返回个数
}
int main()
{
    cin >> x >> y ;
    for (int i = 1 ; i <= x ; i ++)
    {
        ans += num(i) ;//枚举，累加个数
    }
    cout << ans ;
    return 0 ;
}

Java版
import java.util.Scanner;

public class Main
{
public static void main(String[] args)
{
Scanner scan=new Scanner(System.in);
String str=scan.nextLine();
String s=" ";
String[] s1=str.split(s);
int n=Integer.valueOf(s1[0]);
int x=Integer.valueOf(s1[1]);
int num=0;
for(int i=1;i<=n;i++)
{
String ss=String.valueOf(i);
char[] ch=ss.toCharArray();
for(int j=0;j<ch.length;j++)
{
String s2=String.valueOf(ch[j]);
int a=Integer.valueOf(s2);
if(a==x)
{
num++;
}
}
}
System.out.println(num);
}

}

搜索x==0为止 之前每次搜索到一个满足的数字 ans ++
注意ans要开在函数前面
下面是代码

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

import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main {
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int x = scanner.nextInt();
String j = String.valueOf(x);
Pattern pa = Pattern.compile(j);
int count = 0;
for(int i=1;i<=n;i++){
String str = String.valueOf(i);
Matcher ma = pa.matcher(str);
while(ma.find()){
count ++;
}
}
System.out.println(count);
scanner.close();
}

}

import java.io.*;
import java.util.*;

public class Main
{
public static void main(String[] args) throws IOException {
Main main = new Main();
System.out.println(main.getNs(main.x, main.n));
}

private int x;
private int n;

public Main(){
Scanner sc = new Scanner(System.in);
x = sc.nextInt();
n = sc.nextInt();
}

public int getNs(int x, int n){
int counts;
int sum = 0;
int num = 0;
int mod = 0;

for(int i = x ; i > 0 ; i--){
num = i;
counts = 0;
while(num > 0){
mod = num % 10;
num /= 10;
if(mod == n ){
counts++;
}
}
sum += counts;
}
return sum;
}
}

数学小方法

#include<cstdio>
#include<cmath>
int main(){
    int n,x,b=7,o,k;
    scanf("%d%d",&n,&x);
    while(n<=pow(10,--b));
    o=b*pow(10,b-1);
    for(k=pow(10,b);k<=n;k++){
        int j=k;
        while(j){
            if(j%10==x)o++;
            j/=10;
        }
    }
    if(x==0)while(b--)o-=pow(10,b);
    printf("%d",o);
    return 0;
}

水题使人快乐.

#include <iostream>
using namespace std;
int finder(int num,int target){
    int timer=0;
    while(num>=10){
        int x=num%10;
        if(x==target)
            timer++;
        num=(num-x)/10;
    }
    if(num==target)
        timer++;
    return timer;
}
int main(){
    int n,target,i,ans=0;
    scanf("%d%d",&n,&target);
    for(i=1;i<=n;i++)
        ans+=finder(i,target);
    printf("%d",ans);
    return 0;
} 

差点被0关了

import java.util.Scanner;

public class Main {
    static int n = 0;
    static int choice = 0;

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int max = scan.nextInt();
        choice = scan.nextInt();
        findTime(max);
        System.out.println(n);
        scan.close();                                                                       
    }

    public static void findTime(int max) {
        if (max >= choice && choice > 0) {
            n++;
        }
        for (int i = choice + 1; i <= max; i++) {
            findNum(i);
        }
    }

    public static void findNum(int num) {
        if (num < 10) {
            if (num == choice) {
                n++;
            }
            return;
        } else {
            int mod = num % 10;
            if (mod == choice) {
                n++;
            }
            findNum(num / 10);
        }
    }
}

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

#include<stdio.h>
int main(void){
int n,x,i=0,j=0,m,yu=0,end=0,a[50]={0};
scanf("%d %d",&n,&x);
for(i=1;i<=n;i++){
j=0; //reset j
yu=i;
while(yu>=10){
a[j]=yu%10; //循环用来求数的每一位上的数值
yu=yu/10;
j++;
}
a[j]=yu;
m=j;
for(j=0;j<=m;j++){
if(a[j]==x){
end++;
}
}
}
printf("%d\n",end);
return 0;
}

#1 Accepted 103ms 11.305 MiB
#2 Accepted 87ms 9.141 MiB
#3 Accepted 122ms 9.23 MiB
#4 Accepted 106ms 11.055 MiB
#5 Accepted 158ms 8.969 MiB
#6 Accepted 119ms 9.223 MiB
#7 Accepted 91ms 11.328 MiB
#8 Accepted 123ms 11.051 MiB
#9 Accepted 107ms 9.328 MiB

#10 Accepted 148ms 9.121 MiB

using System;

namespace ConsoleApplication2
{
class Program
{
static void Main(string[] args)
{
string str = Console.ReadLine();
string[] sn = str.Split(' ');
int n = int.Parse(sn[0]);
int x = int.Parse(sn[1]);
int count = 0;
for (int i = 1; i <=n; i++)
{
int num = i;
while(num>0)
{
if (num % 10 == x)
count++;
num/=10;
}
}
Console.WriteLine(count);
}
}
}

//一道简单的取数判断题

#include <iostream>
using namespace std;

int main()
{
long int n;//定义一个n，作为要输入的1到n
int x,i=1,a,b=0,z=0;
cin>>n>>x;
for(;i<=n;i++)
{a=i;//把外循环的1-n次每个数 都赋给a然后进行判断
while(a)
{ b=a%10;//a的余数的值赋给b，意思就是取个位数
if(b==x)//进行判断若b是否等于所输入的x，
z++;//满足，则计数器z
a=a/10;
}//然后让a/10继续进行判断直到a=0结束while循环
//然后跳回外循环进行第二个数的判断

}

cout<<z<<endl;

return 0;
}

//一道简单的取数判断题

#include <iostream>
using namespace std;

int main()
{
long int n;//定义一个n，作为要输入的1到n
int x,i=1,a,b=0,z=0;
cin>>n>>x;
for(;i<=n;i++)
{a=i;//把外循环的1-n次每个数 都赋给a然后进行判断
while(a)
{ b=a%10;//a的余数的值赋给b，意思就是取个位数
if(b==x)//进行判断若b是否等于所输入的x，
z++;//满足，则计数器z
a=a/10;
}//然后让a/10继续进行判断直到a=0结束while循环
//然后跳回外循环进行第二个数的判断

}

cout<<z<<endl;

return 0;
}

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

思路很简单，直接求出1~n每个数的x出现次数。。。
#include<bits/stdc++.h>
using namespace std;
int n,x,ans;
void check(int z){
int a[15],zx=z,p=-1;
while(zx!=0){
p++;
a[p]=zx%10;
zx/=10;
}
for(int i=0;i<=p;i++)
if(a[i]==x) ans++;
return;
}
int main(){
scanf("%d%d",&n,&x);
for(int i=1;i<=n;i++) check(i);
printf("%d\n",ans);
return 0;
}

n, x = raw_input().split()
print sum([str(i+1).count(x) for i in range(int(n))])

int a[10];
int main(){
    int n,x;
    scanf("%d%d", &n, &x);
    for(int i=1; i<=n; i++){
        int k=i;
        while(k){
            a[k%10]++;
            k/=10;
        }
    }
    printf("%d", a[x]);
    return 0;
}

Haskell到达战场

main=print.(\(n:x:[])->sum.map(length.filter(\n->read[n]==x).show)$[1..n]).map read.words=<<getLine

可惜上面这种会超时，只能改成下面这个：

solve :: Int -> Int -> Int
solve 0 _   = 0
solve num x = number num x + solve (num - 1) x

number :: Int -> Int -> Int
number 0 _ = 0
number num x
  | num `mod` 10 == x = 1 + number (num `quot` 10) x
  | otherwise         = number (num `quot` 10) x

main :: IO()
main = print . (\(a:b:[])->solve a b) . map read . words =<< getLine