题解 | 躲藏
躲藏
https://www.nowcoder.com/practice/b19eda9490d54505a1065a70c3005b47
/*
* @Author: tkzzzzzz6
* @Date: 2026-02-02 21:39:02
* @LastEditors: tkzzzzzz6
* @LastEditTime: 2026-02-02 21:55:31
*/
#include<bits/stdc++.h>
#define il inline
#define endl '\n'
using namespace std;
#define pb push_back
#define fastio \
ios::sync_with_stdio(false); \
cin.tie(0);
typedef long long ll;
typedef unsigned long long ull;
const ull MOD = 2000120420010122;
il void solve(){
string s;
while(getline(cin,s)){
ull dp[4] = {0,0,0,0};
// dp[0] 'c'的方案数
// dp[1] 'cw'的方案数
// dp[2] 'cwb'的方案数
// dp[3] 'cwbc'的方案数
const ull base = 1;
for(unsigned char ch:s){
char x = (char)tolower(ch);
if(x == 'c'){
dp[3] += dp[2];
if(dp[3] >= MOD){
dp[3] %= MOD;
}
// 对于开头'c'的方案处理
dp[0] += base;
if(dp[0] >= MOD){
dp[0] %= MOD;
}
}else if(x == 'w'){
dp[1] += dp[0];
if(dp[1] >= MOD){
dp[1] %= MOD;
}
}else if(x == 'b'){
dp[2] += dp[1];
if(dp[2]>=MOD){
dp[2] %= MOD;
}
}
}
cout << dp[3] % MOD << endl;
}
}
int main()
{
fastio;
int t = 1;
// cin >> t;
while(t--)
{
solve();
}
return 0;
}
