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 positioni” 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 1000000007int 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;elsedp[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);}