[CF1303E] Erase Subsequences - dp】的更多相关文章

Solution 不由分说地枚举分割点 令 \(f[i][j]\) 表示原串处理到 \(i\) ,\(s_1\) 处理到 \(j\),\(s_2\) 最多能处理到哪里 采用主动转移 任意情况, \(f[i][j] \to f[i+1][j]\) 如果 \(s[i+1]=s_1[j+1]\) ,那么 \(f[i][j] \to f[i+1][j+1]\) 如果 \(s[i+1]=s_2[f[i][j]+1]\) ,那么 \(f[i][j]+1 \to f[i+1][j]\) 时间复杂度 \(O(n…
https://codeforces.com/contest/1303/problem/E #include<bits/stdc++.h> using namespace std; ; int dp[maxn][maxn]; bool check(string s,string t){ ; ;i<s.length();i++){ if(t[indx] == s[i]) indx++; } if(indx == t.length()) return true; return false;…
题目链接: C. Subsequences time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output For the given sequence with n different elements find the number of increasing subsequences with k + 1 elements. It is…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2227 Find the nondecreasing subsequences                                  Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)                                             …
Problem Description How many nondecreasing subsequences can you find in the sequence S = {s1, s2, s3, ...., sn} ? For example, we assume that S = {1, 2, 3}, and you can find seven nondecreasing subsequences, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1,…
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative…
http://acm.hdu.edu.cn/showproblem.php?pid=2227 用dp[i]表示以第i个数为结尾的nondecreasing串有多少个. 那么对于每个a[i] 要去找 <= a[i]的数字那些位置,加上他们的dp值即可. 可以用树状数组维护 #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algori…
转换一下题意,就相当于问t能不能和s中2个不相同的子串相同,我们可以将t串拆成2个子串t1,t2,得到状态dp[i][j][k]=0/1,表示s判断到i位,t1判断到j位,t2判断到k位,0/1表示是否满足 两个状态,s[i]与t1[j]相匹配,s[i]与t2[k]相匹配 dp[i+1][j+1][k] = dp[i][j][k] dp[i+1][j][k+1] = dp[i][j][k] 因为dp的值只取0和1,我们可以优化dp函数,dp[i][j][k]可以变成dp[i][j] = k,这样…
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A. Erasing Zeroes (模拟) #include<bits/stdc++.h> using namespace std; typedef long long ll; ; int main(){ int t;cin>>t; while(t--){ string s;cin>>s; ,f2 = ; ; ; ;i<s.length();i++){ '){ ) { ans+=cnt; cnt = ; f1 = ; } ) f1 = ; } else{ ) c…