#include<iostream>
using namespace std;
int main()//水题+动态数组
{
    int min,max;cin>>min>>max;
    int *a;
    a=new int [10001];
    int cnt=0;
    for(int i=min;i<=max;i++)
    {
        int j=i;
        int d;
        while(j>0)
        {
            d=j%10;
            j/=10;
            if(d==2) cnt++;
            d=0;
        }
    }   
    cout<<cnt;
    return 0;
}

var
n,t,l,k,num,i:longint;
s,num1:string;
begin
read(n,num);
str(2,num1);
for i:=n to num 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.
哎呀，空大了！！！

#include <stdio.h>
int a[10];
int main(){
    int n,m;
    scanf("%d%d", &m, &n);
    for(int i=m; i<=n; i++){
        int k=i;
        while(k){
            a[k%10]++;
            k/=10;
        }
    }
    printf("%d", a[2]);
    return 0;
}

水题不解释

#include<iostream>
using namespace std;
int l,r,ans;
int main()
{
    cin>>l>>r;
    for(int i=l;i<=r;i++)
    {
        int a=i;
        while(a>0)
        {
            if(a%10==2) ans++;
            a/=10;
        }
    }
    cout<<ans;
    return 0;
}

纯水

#include<iostream>
using namespace std;
int main()
{
    int l,r,i,ans=0,x;
    cin>>l>>r;
    for(i=l;i<=r;i++)
    {
        x=i;
        while(x>0)
        {
            if(x%10==2)
            ans++;
            x/=10;
        }
    }
    cout<<ans;
    return 0;
} 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<math.h>

using namespace std;

vector<int> w;

void f(int a){
    while (a>0){
        w.push_back(a%10);
        a/=10;
    }
}

int main(void){
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    int l,r;
    cin>>l>>r;
    for (int i=l;i<=r;i++){
        f(i);
    }
    int ans=0;
    for (int i=0;i<w.size();i++){
        if (w[i]==2) ans++;
    }
    cout<<ans;
}

waterful

so water 暴搜即可！！！！！！！！！！！！！！！！！

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
    long long n,all=0,a,i;
    scanf("%lld %lld",&a,&n);
    for(i=a;i<=n;i++)
    {
        long long aha=i;
        while(aha>=1)
        {
            if(aha%10==2)
                all++;
            aha=aha/10;
        }
    }
    printf("%lld",all);
    return 0;
}

题目很简单，方法步骤如下：
1. 输入两个数（范围）
2. 创造循环每个数，比较有没有“2”
3. 把“2”的数量算出来

注意：
* 因为题目要求是范围在10000,（1≤L≤R≤10000）
* 所以只要比较到万位
* 将每个位的数进行判断，凡有一个次数就加1

代码如下

#include<stdio.h>
int main()
{
    int sum=0;
    int a;
    int b;
    int x;
    scanf("%d %d",&a,&b);
    for(x=a;x<=b;x++)
    {
        if((x/1)%10==2)//个位 
        sum++;
        if((x/10)%10==2)//十位 
        sum++;
        if((x/100)%10==2)//百位
        sum++;
        if((x/1000)%10==2)//千位 
        sum++;
        if((x/10000)%10==2)//万位 
        sum++;
    }
    printf("%d",sum); 
} 

#include<iostream>
using namespace std;
int main()
{
    long long a,b,s=0,n;
    cin>>a>>b;
    for(int y=a;y<=b;y++)
    {
        n=y;
        while(n!=0)
        {
            if(n%10==2) s++;
            n/=10;
        }
    }
    cout<<s;
        return 0;
}

直接字符吧
#include<bits/stdc++.h>
using namespace std;
char s[10];
int main()
{

int l,r,ans=0;
cin>>l>>r;
for(int i=l;i<=r;i++)
{
sprintf(s,"%d",i);
l=strlen(s);
for(int j=0;j<=l-1;j++)
if(s[j]=='2')
ans++;
}
cout<<ans;
}

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int l,r,a,num=0;
cin>>l>>r;
for (int i=l;i<=r;i++)
{
if (2==i/10000) num+=1;
if (2==(i%10000)/1000) num+=1;
if (2==(i%1000)/100) num+=1;
if (2==(i%100)/10) num+=1;
if (2==i%10) num+=1;
}
cout<<num;
return 0;
}

自定义函数
#include<bits/stdc++.h>
using namespace std;
int js(int n){
int x,s=0;
while(n>=1){
x=n;
x=x%10;
n=n/10;
if(x==2)
s++;
}
return s;
}
int main(){
int l,r,s=0,i;
cin>>l>>r;
for(i=l;i<=r;i++)

s+=js(i);
cout<<s;
return 0;
}

三分钟AC，纯属练手

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

int a,b;

int main()
{
    scanf("%d%d",&a,&b);
    int ans=0,u,bt;
    for(int i=a;i<=b;++i)
    {
        u=i;
        while(u>0)
        {
            bt=u%10;
            if(bt==2)ans++;
            u=(u-bt)/10;
        }
    }
    printf("%d",ans);
    return 0;
} 

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

int main(){
int a,b=0,c,d;
cin>>a>>c;
for(int i=a;i<=c;i++){
d=i;
while(d!=0){
if(d%10==2){
b++;
}
d=d/10;
}
}
cout<<b<<endl;
}

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

int fj(int x)
{
int ans=0,y=x;
while(y)
{
if(y%10==2)
ans++;
y/=10;
}
return ans;
}

int main()
{
//freopen("数字统计.in","r",stdin);
//freopen("数字统计.out","w",stdout);
int a,b,ans=0;
scanf("%d%d",&a,&b);
for(int i=a;i<=b;i++)
ans+=fj(i);
printf("%d\n",ans);
return 0;
}

#include <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
int main()
{
int l,r,x,s=0;
scanf("%d%d",&l,&r);
for(int i=l;i<=r;i++)
{
x=i;
while(x>0)
{
if(x%10==2)
{
s++;
}
x=x/10;
}
}
printf("%d\n",s);
return 0;
}

算法很简单，只需要判断这两个数范围内每个数，将范围内每个数的各个位的数分离出来，如果分离出来的各个数中有一个数为2，就使计数器加一，最后输出这个计数器就好。

#include<stdio.h>
int main(){
int L,R;
scanf("%d %d",&L,&R);
int n=0;
for(;L<=R;L++)
{
if((L/1000)==2)
n++;
if((((L/100)>10)?((L/100)-10*(L/1000)):(L/100))==2)
n++;
if((((L/10)>10)?((L/10)-10*(L/100)):(L/10))==2)
n++;
if(L%10==2)
n++;
}

printf("%d",n);
return 0;
}

水数据！水题！

var n,m,i,j,h:longint;
a:array[1..100000] of string;
begin
read(n,m);
for i:=n to m do
str(i,a[i]);
for i:=1 to m do
if a[i]='' then a[i]:='LYY';
i:=0;
repeat
inc(i);
for j:=1 to 20 do
if a[i][j]='2' then inc (h);
until a[i]='';
write(h);
end.

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