视频讲解
代码
// idx 0 1 2 3 4 5
// S = A A B B A C
// f[i][u][j]
// u = {A,C}, {A}
// A B C
// 1 0 1
// not choose S[i]
// f[i][u][j] += f[i-1][u][j]
// choose S[i] = x
// 连续
// j == x(S[i])
// f[i][u][j] += f[i-1][u][j]
// 不连续
// v' = x + v
// j != x
// f[i][v'][x] += f[i-1][v][j]
// f[i][W][x] = 1 W 中只包含 x 的情况下
include
using namespace std;
long long f[1024][1024][10];
define mod 998244353
int n;
string S;
int main() {
cin>>n;
cin>>S;
memset(f, 0, sizeof f);
for (int i=1;i<=n;++i) {
auto x = S[i-1] - 'A';
// not choose S[i]
for (int u=0;u<1024;++u) {
for (int j=0;j<10;++j) {
f[i][u][j] += f[i-1][u][j];
f[i][u][j] %= mod;
if (x == j) {
// choose S[i] = x 连续
f[i][u][j] += f[i-1][u][j];
f[i][u][j] %= mod;
}
}
}
// choose S[i] = x 不连续
for (int v=0;v<1024;++v) {
if (v&(1<<x)) continue;
for (int j=0;j<10;++j) {
f[i][v|(1<<x)][x] += f[i-1][v][j];
f[i][v|(1<<x)][x] %= mod;
}
}
// default
f[i][1<<x][x]++;
}
long long ans = 0;
for (int u=0;u<1024;++u) {
for (int j=0;j<10;++j) {
ans += f[n][u][j];
ans %= mod;
}
}
cout<<ans<<endl;
return 0;
}参考
[1]. AtCoder ABC 215 E. https://atcoder.jp/contests/abc215/tasks/abc215_e
欢迎分享，引用。若转载，请联系我，谢谢配合。 本文地址：https://qoogle.top/xiaoxu-tutorial-bas-bitdp-chain-contestant/
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