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用f[i][j]表示a串计算到i位，b串计算到j位时的最长公共子序列长度

g[i][j]表示a串计算到i位，b串计算到j位时的最长公共子序列个数

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn=5000+5;
const int mod=1e8;
char s1[maxn],s2[maxn];
int len1,len2,f[maxn][maxn];
long long g[maxn][maxn];
int main()
{freopen("lcs.in","r",stdin);freopen("lcs.out","w",stdout);char cc;for(;;){scanf("%c",&cc);if(cc=='.') break;s1[++len1]=cc;}for(;;){scanf("%c",&cc);if(cc=='.') break;s2[++len2]=cc;}for(int i=0;i<=len1;i++) g[i][0]=1;for(int i=0;i<=len2;i++) g[0][i]=1;for(int i=1;i<=len1;i++)for(int j=1;j<=len2;j++){if(s1[i]==s2[j]) {f[i][j]=max(f[i][j],f[i-1][j-1]+1);f[i][j]=max(f[i][j],f[i-1][j]);f[i][j]=max(f[i][j],f[i][j-1]);if(f[i][j]==f[i-1][j-1]+1) g[i][j]+=g[i-1][j-1],g[i][j]%=mod;if(f[i-1][j]==f[i][j]) g[i][j]+=g[i-1][j],g[i][j]%=mod;if(f[i][j-1]==f[i][j]) g[i][j]+=g[i][j-1],g[i][j]%=mod;}else{f[i][j]=max(f[i][j],f[i-1][j]);f[i][j]=max(f[i][j],f[i][j-1]);if(f[i-1][j]==f[i][j]) g[i][j]+=g[i-1][j],g[i][j]%=mod;if(f[i][j-1]==f[i][j]) g[i][j]+=g[i][j-1],g[i][j]%=mod;if(f[i-1][j-1]==f[i][j]) g[i][j]-=g[i-1][j-1],g[i][j]%=mod;}}printf("%d\n%lld",f[len1][len2],(g[len1][len2]+mod)%mod);return 0;
} 

按照题解给的实现会MLE，然后把它滚动成一维的就能A了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn=5000+5;
const int mod=1e8;
char s1[maxn],s2[maxn];
int len1,len2,f[2][maxn];
long long g[2][maxn];
int main()
{freopen("lcs.in","r",stdin);freopen("lcs.out","w",stdout);char cc;for(;;){scanf("%c",&cc);if(cc=='.') break;s1[++len1]=cc;}for(;;){scanf("%c",&cc);if(cc=='.') break;s2[++len2]=cc;}for(int i=0;i<=len2;i++) g[0][i]=1;int i,ii; g[0][0]=1;for(ii=1,i=1;ii<=len1;ii++,i=i^1){memset(g[i],0,sizeof(g[i]));g[i][0]=1;for(int j=1;j<=len2;j++){if(s1[ii]==s2[j]) {f[i][j]=0;f[i][j]=max(f[i][j],f[i^1][j-1]+1);f[i][j]=max(f[i][j],f[i^1][j]);f[i][j]=max(f[i][j],f[i][j-1]);if(f[i][j]==f[i^1][j-1]+1) g[i][j]+=g[i^1][j-1],g[i][j]%=mod;if(f[i^1][j]==f[i][j]) g[i][j]+=g[i^1][j],g[i][j]%=mod;if(f[i][j-1]==f[i][j]) g[i][j]+=g[i][j-1],g[i][j]%=mod;}else{f[i][j]=0;f[i][j]=max(f[i][j],f[i^1][j]);f[i][j]=max(f[i][j],f[i][j-1]);if(f[i^1][j]==f[i][j]) g[i][j]+=g[i^1][j],g[i][j]%=mod;if(f[i][j-1]==f[i][j]) g[i][j]+=g[i][j-1],g[i][j]%=mod;if(f[i^1][j-1]==f[i][j]) g[i][j]-=g[i^1][j-1],g[i][j]%=mod;}}}printf("%d\n%lld",f[i^1][len2],(g[i^1][len2]+mod)%mod);return 0;
}