#include<bits/stdc++.h>
using namespace std;
string waj(int i){
string temp;
ostringstream oss;
oss<<i;
temp=oss.str();
return temp;
}
int main(){
int l,r,ans=0;
cin>>l>>r;
string a;
for(int i=l;i<=r;i++){
a+=waj(i);
}
for(int i=0;i<a.size();i++){
if(a[i]=='2') ans++;
}
cout<<ans;
}

###**dp**

评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 548 KiB, score = 10
Accepted, time = 0 ms, mem = 552 KiB, score = 100
代码
#include <cstdio>
int dp[10001],L,R,sum = 0;
int main() {
  scanf("%d%d",&L,&R);
  dp[2] = 1;
  for (int i = 1;i <= R;i++) {
    dp[i] = dp[i%10];
    if (i/10) dp[i] += dp[i/10];
    if (i >= L) sum += dp[i];
  }
  printf("%d",sum);
  return 0;
}

逐位判断竟然也能0ms。。。

字符串 打遍天下无敌手
pascal
var
l,r,i,s,j:longint;
s1:string;
begin
readln(l,r);
for i:=l to r do
begin
str(i,s1);
for j:=1 to length(s1) do
if s1[j]='2'
then
inc(s);
end;
writeln(s);
readln;
end.`

var n,m,i,s,j:longint;
st:string;
begin
readln(n,m);
for i:=n to m do
begin
str(i,st);
for j:=1 to length(st) do
if st[j]='2' then inc(s);
end;
writeln(s);
end.

import java.util.Scanner;
public class Main {

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        String str;
        StringBuffer sb=new StringBuffer();
        int begin=sc.nextInt();
        int end=sc.nextInt();
        for(int i=begin;i<=end;i++)
            sb.append(i);
        str=sb.toString();
        System.out.println(countToken(str, "2"));
    }
    public static int countToken(String str,String token){
        int count=0;
        while(str.indexOf(token)!=-1){
        count++;
        str = str.substring(str.indexOf(token)+token.length());
        }
        return count;
        }

}

#include<cstdio>
using namespace std;

int main()
{
int r, l, ans = 0;
scanf("%d%d", &r, &l);
for(int i=r; i<=l; i++){
int bri = i;
while(bri > 0){
if(bri%10 == 2)
ans++;
bri /= 10;
}
}

printf("%d", ans);
return 0;
}
好水

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
long long l,r;
long long s=0;
cin>>l>>r;
char a[1000];
for(int i=l;i<=r;i++)
{
a[i]=i;
while(a[i]>0)
{
if(a[i]%10==2)
++s;
a[i]=a[i]/10;
}
}
cout<<s;
return 0;
} 哪里错了

原来不用string判断也是那么简单...

Block code

#include <iostream>
#include <stdio.h>
int make2(int v) {
if (v > 10) {
int p = v % 10;
return make2(p) + make2((v - p) / 10);
}
if (v == 2)
return 1;
return 0;
}
int main(int argc, char **argv)
{
int L; int R; int val = 0;
std::cin>>L; std::cin>>R;
for (;L <= R;L++)
val += make2(L);
std::cout<<val;
return 0;
}

var b: string;
a,i,j,l,r,tri:longint;
procedure two(l,r:longint);
begin
for i:=l to r do
begin
str(i,b);
for j:=1 to length(b) do
if b[j]='2' then inc(tri);
end;
end;
begin
readln(l,r);
two(l,r);
write(tri);
end.

#include<stdio.h>
#include<string.h>
char ch[100];
int main()
{
int i,l,r,ans=0,x;
scanf("%d%d",&l,&r);
for(i=l;i<=r;i++)
{
x=i;
while(x>0)
{
if(x%10==2)
ans++;
x/=10;
}
}
printf("%d",ans);
return 0;
}

Pascal Code

var
l,r:longint;
i,n,ans,t:longint;
begin
readln(l,r);
for i:=l to r do
begin
t:=i;
repeat //取出每一位
n:=t mod 10;
if n=2 then inc(ans);
t:=t div 10;
until t=0;
end;
writeln(ans);
end.

秒过。。。
大水题。。。
var l,r,i,ans:integer;
function num(k:integer):integer;
var t:byte;
begin
num:=0;
repeat
t:=k mod 10;
if t=2 then inc(num);
k:=k div 10;
until k=0;
end;
begin
readln(l,r);
for i:=l to r do
inc(ans,num(i));
writeln(ans);
end.

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int i,l,r,num=0;
cin>>l>>r;
for(i=l;i<=r;i++)
{
int t,x;
t=i;
while (t>0)
{
x=t % 10;
if (x==2)
num+=1;
t=t/10;

}
}
cout<<num;
}

略水……不过……

AC 100 留念

###Blockcode

var
sum,l,r,i,j,p:longint;
num:string;
begin
sum:=0;
read(l);
read(r);
for i:=l to r do
begin
str(i,num);
p:=pos('2',num);
while p>0 do
begin
inc(sum);
delete(num,p,1);
p:=pos('2',num);
end;
end;
write(sum);
end.

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
main()
{
char s[10];
int a=0,e,i,j,l,r;
cin>>l>>r;
for(i=l;i<=r;i++)
{
sprintf(s,"%d",i);
e=strlen(s);
for(j=0;j<e;j++)
if(s[j]=='2')
a++;
}
cout<<a;
}

program p1784;
var sum,l,r,i,j:longint;
num:string;
begin
sum:=0;
read(l); read(r);
for i:=l to r do
begin
str(i,num);
for j:=1 to length(num) do
if num[j]='2' then
sum:=sum+1;
end;
write(sum);
end.

简单易懂，自己看吧。如果这题数据规模大点貌似就有点贪心的味道了

C++ Code (试试stringstream)

#include <iostream>
#include <string>
#include <sstream>

using namespace std;

string str;

int main(void)
{
stringstream ss(str);
long long ans(0);
int n(0), m(0);
cin >> n >> m;
for (int i = n; i <= m; ++i)
{
ss << i;
}
ss >> str;
for (int i = 0; i < str.length(); ++i)
{
if (str[i] == '2') { ++ans; }
}
cout << ans << endl;
return 0;
}

C++ Code

#include<iostream>
#include<string>
using namespace std;
int check(int i);
int main()
{
int start,end,i,count;
count=0;
cin>>start>>end;
for(i=start;i<=end;i++)
{
count+=check(i);
}
cout<<count;
return 0;
}
int check(int i)
{
int temp,count;
count=0;
do
{
temp=i%10;
i=i/10;
if(temp==2) count++;
}while(i!=0) ;
return count;
}

var
l,r:longint;
i,sum,n:longint;
begin
sum:=0;
read(l,r);
for i:=l to r do
begin
n:=i;
while n<>0 do
begin
if n mod 10=2 then sum:=sum+1;
n:=n div 10;
end;
end;
writeln(sum);
end.