字符串中子序列出现次数【模板题】

形如：题意：给一个字符串，求子序列“xyz...x”出现的次数

两道板子题：https://ac.nowcoder.com/acm/contest/3947/I

 https://ac.nowcoder.com/acm/contest/87/B【注意：是否区分大小写】

#include<bits/stdc++.h>

using namespace std;
#define int long long
#define mod 20010905 
int dp[66];
signed main(){
    string s;
    cin>>s;
    
    for(int i=0;i<s.size();i++){
        s[i]=tolower(s[i]);
        dp[8]=(dp[8]+(s[i]=='u')*dp[7])%mod;
        dp[7]=(dp[7]+(s[i]=='o')*dp[6])%mod;
        dp[6]=(dp[6]+(s[i]=='y')*dp[5])%mod;
        dp[5]=(dp[5]+(s[i]=='e')*dp[4])%mod;
        dp[4]=(dp[4]+(s[i]=='v')*dp[3])%mod;
        dp[3]=(dp[3]+(s[i]=='o')*dp[2])%mod;
        dp[2]=(dp[2]+(s[i]=='l')*dp[1])%mod;
        dp[1]=(dp[1]+(s[i]=='i'))%mod;
    }
    cout<<dp[8];
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int inf=1<<30;
typedef long long ll;
const double pi=acos(-1);
const int mod=2000120420010122;
const int maxn=2e5+7;
ll dp[5];
int main(){
    string s;
    while(cin>>s){
    memset(dp,0,sizeof(dp));
    for(int i=0;i<s.length();i++){
        s[i]=tolower(s[i]);            
            dp[4]=(dp[4]+(s[i]=='c')*dp[3])%mod;
            dp[3]=(dp[3]+(s[i]=='b')*dp[2])%mod;
            dp[2]=(dp[2]+(s[i]=='w')*dp[1])%mod;
            dp[1]=(dp[1]+(s[i]=='c'))%mod;            
    }
    cout<<dp[4]<<endl;
    }
    return 0;
}