Codeforces 164B || Codeforces 163A

Codeforces 164B

是个字符串问题（codeforces把这类问题归为two pointers），给了2个字符串a,b都可以任意移位，且b中无相等元素，满足条件“是a的子串==是b的子序列”的最大的串

#include<cstdio> #include<cstring> #include<algorithm> #include<queue> using namespace std; int p[1000010],a[1000010],n,m,ans; deque<int>q; bool in(int x,int L,int R){ if(L<=R)return L<=x&&x<=R; else return x<=R||L<=x; } int main(){ scanf("%d%d",&n,&m); for(int i=0;i<n;++i)scanf("%d",&a[i]); memset(p,-1,sizeof p); for(int i=0,x;i<m;++i){ scanf("%d",&x); p[x]=i; // 利用串里无相同元素 } for(int i=0;i<n+n;++i){ if(p[a[i%n]]==-1){ q.clear(); continue; } while(!q.empty()&&in(p[a[i%n]],q.front(),q.back())) //in函数判断p[a[i%n]]必须在选出的b序列的外面，否则的话队首元素出队 q.pop_front(); q.push_back(p[a[i%n]]); ans=max(ans,(int)q.size()); if(ans>=min(m,n)) break; } printf("%d\n",ans); return 0; }

 Codeforces 163A

（x,y）其中x==y,x为字符串a的子串，y为字符串b的子序列，问这样的序偶有多少个？

解：这是个DP的问题

The problem could be solved with the following dynamic programming. Let f[i, j] be the number of distinct pairs (“substring starting at position i” and “subsequence of the substring t[j… |t|]”)

Then:

f[i, j] = f[i, j + 1];
if (s[i] == t[j])
  add(f[i, j], f[i + 1, j + 1] + 1)

Answer = f[i,0]

#include<cstdio> #include<cstring> #include<algorithm> #include<queue> using namespace std; #define mod 1000000007 int dp[5010][5010]; int main(){ int i,j; char s1[5010],s2[5010]; scanf("%s%s",s1,s2); int len1=strlen(s1),len2=strlen(s2); for(i=len1-1;i>=0;i--) for(j=len2-1;j>=0;j--){ if(s1[i]==s2[j]) dp[i][j]=(dp[i][j+1]+1+dp[i+1][j+1])%mod; else dp[i][j]=(dp[i][j+1])%mod; } int sum=0; for(i=len1-1;i>=0;i--) sum=(sum+dp[i][0])%mod; printf("%d\n",sum); }