A.

```/*
It is found that each reversal or elimination will reduce a period of time 0
When 0 It can only be eliminated for a period of time
This is the way to judge
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<set>
#include<map>
#define M 300010
#define ll long long
using namespace std;
int nm = , f = ;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -;
for(; isdigit(c); c = getchar()) nm = nm * + c - '';
return nm * f;
}
ll n,x,y;
char s[M];
int main() {
scanf("%s", s + );
int len = strlen(s + );
s[] = '?';
ll tot = ;
for(int i = ; i <= len; i++) if(s[i] != s[i - ] && s[i] == '') tot++;
if(tot == ) return puts("");
cout << min(tot * y, tot * x - x + y);
return ;
}```

B.

```/*
Maybe this kind of question is a table killer
There is no rule in the small data section Big data, regular
// An obvious conclusion , If the total number is certain, we can find 1, 5, 10, 50 The number of different sums added up is equivalent to finding 0, 4, 9, 49 Of
Well, it's time to hit the clock
For the former 12 The data of Direct violence , The linear increase in the back
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<set>
#include<map>
#define M 30
#define ll long long
using namespace std;
int nm = , f = ;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -;
for(; isdigit(c); c = getchar()) nm = nm * + c - '';
return nm * f;
}
const ll dx[]={,,,,,,,,,,,,,,,};
int main() {
if(n <= ) cout << dx[n];
else cout << dx[] + (n - ) * ;
return ;
}```

C.

Obviously, a line of the same color contributes the same to the answer in which line , So we can get the formula of inclusion and exclusion directly

```/*
difficult Look at the solution
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<set>
#include<map>
#define M 1231231
#define ll long long
const int mod = ;
using namespace std;
int nm = , f = ;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -;
for(; isdigit(c); c = getchar()) nm = nm * + c - '';
return nm * f;
}
ll poww(ll a, ll b) {
ll as = , tmp = a;
for(; b; b >>= , tmp = tmp * tmp % mod) if(b & ) as = as * tmp % mod;
return as;
}
ll c[M];
inline ll ni(ll a) {
return poww(a, mod - );
}
void shai(ll n) {
c[] = ;
for(int i = ; i <= n; i++) c[i] = c[i - ] * (n - i + ) % mod * ni(i) % mod;
}
int main() {
ll ans = ;
shai(n);
for(int i = , j = ; i <= n; i++, j = -j) ans += j * c[i] % mod * poww(, (n - i) * n + i) % mod, ans %= mod;
ans = ans * % mod;
for(int i = , j = -; i < n; i++, j = -j) ans += 3ll * c[i] * j % mod * (poww(1ll - poww(, i), n) - poww(-1ll * poww(3ll, i), n)) % mod, ans %= mod;
cout << (ans + mod) % mod;
return ;
}```

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