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G title: subsequence

The meaning of problems: Given a two-character string numbers s, t, is calculated as the number of the sub-sequence s after a positive integer larger than the sub-sequence of t.

Analysis: subsequence length t s is greater than a certain ratio of large t (negative beginning with 0), by combining a number of pre-treatment can be obtained, the key operator sequence length equal to t s is greater than t, there are several. Clearly use herein dp practice, with s, t from right to left in a state of the position mark dp, dp [i] [j] indicates the state: the maximum number of strings before i and s j front string of t . For each state, it must contain a s position on the state, then according to a new first entry is the same as the first item plus t more than the number, if greater than , to use selected principles are selected on plus a number of combinations of the selected head, if the same can also add value dp are recorded on an ever selected.

#include<cstdio>
using namespace std;

typedef long long ll;
const int maxn=3007;
const int mod=998244353; 
ll C[maxn][maxn];
ll dp[maxn][maxn];
char s[maxn],t[maxn];

void init_C(){
    C[0][0]=1;
    for(int i=1; i<maxn; i++){
        C[i][0]=1;
        for(int j=1; j<=maxn; j++){
            C[i][j] = ( C[i-1][j]+C[i-1][j-1] )%mod;
        }
    }
}
int main(){
    init_C();
    int T;
    scanf("%d",&T);
    int n,m;
    while(T--){
        scanf("%d%d",&n,&m);
        scanf("%s%s",s,t);
        for(int i=0; i<=n; i++){
            for(int j=0; j<=m; j++){
                dp[i][j]=0;
            }
        }
        ll ans=0;
        //线性递推的二维dp
        for(int i=m-1; i>=0; i--){
            for(int j=n-1; j>=0; j--){
                ( dp[j][i] += dp[j+1][i] )%=mod;    //Security at the end of the 
                IF (S [J] == T [I]) (DP [J] [I] + = DP [J + . 1 ] [I + . 1 ])% = MOD;    // can add up opportunities for 
                the else  IF ( S [J]> T [I]) (DP [J] [I] + = C [N- . 1 -j] [M- . 1 -i]);      // selected thought, it is selected which one C dp recursive calculation rather than direct, meaning brutal growth 
            } 
        } 
        ANS + dp = [ 0 ] [ 0 ];
 //         the printf ( "% LLD ANS = \ n-", ANS); 
        
        for ( int I = N- . 1 ; I > = 0 ; i-- ) {
             IF ! (S [I] = ' 0 ') {
                 For ( int J = Ni; J> = m + . 1 ; J, ) {
                     // the printf ( "% D% D \ n-", I, J); 
                    (ANS + C = [N- . 1 -i] [ J- . 1 ]) MOD =%;         // selected principles supra 
                } 
            } 
        } 
        the printf ( " % LLD \ n- " , ANS); 
    }     
}

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