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## 复习提纲

以$A$为例, 5进制理解为$5^0,5^1,5^2,5^3$
$(1234)_5=1*4+3*5+2*25+1*125=194$
$(302)_8=2*8^0+0*8^1+3*8^2=2*1+0*8+3*64=194$
$(11000100)_2=(304)_8$ 与$(302)_8$不同,所以答案是$C$
$(C2)_{16}=2*1+12*16=194$

```cpp {.line-numbers}
#include <bits/stdc++.h>
using namespace std;
bool check(int x) {
int sum = 0;
while (x) {
int a = x % 10;
x /= 10;
sum += a;
}
return sum % 2 == 0;
}
int main() {
int n, m;
cin >> n >> m;
for (int i = n; i < = m; i++)
if (check(i))
cout < < i << endl ;
return 0 ;
}
```

```cpp {.line-numbers}
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
int m;
cin >> m;
m--;
while (m--) {
string t;
for (int i = 0; i < s.size(); i++) {
int c = 1;
while (i + 1 < s.size() && s[i + 1] == s[i]) {
c++;
i++;
}
t = t + to_string(c) + s[i];
}
s = t;
}
cout << s << endl;
return 0;
}
```

``` cpp {. line-numbers }
#include <bits/stdc++.h>
using namespace std ;
int k ;
string kill ( string s ) {
int p = s[s.size() - 1 ]; // 最后一位是默认值
for ( int i = 0; i < s . size () - 1 ; i ++)
if ( s [ i + 1 ] > s[i]) {
p = i;
break;
}
string res;
for (int i = 0; i < s.size (); i ++)
if ( i != p ) res += s [ i ];
return res ;
}
int main () {
string s ;
cin > > s >> k;
for (int i = 1; i < = k; i++) s = kill(s);
cout < < s << endl ;
return 0 ;
}
```