README.md

@@ -1,10 +0,0 @@
-## 线性动态规划优化为$O(n\log{n})$方法
->如果是递增序列就lower_bound
-
->如果是递减序列就手写二分
-
-## 区间dp
-1. 根据问题推出dp含义
-2. 根据规则写出dp的状态转移公式
-3. 处理边界问题
-> dp[i][j], dp[0][0], dp[i][0], dp[0][j], dp[i][i], dp[j][j]

src/8/26/P2758.cpp

@@ -1,53 +0,0 @@
-/*
-
-dp[i][j]=源串前i个字符匹配目标串前j个字符所需要的最小操作次数
-
-如果第i个字符==第j个字符
-dp[i][j]=dp[i-1][j-1]
-如果不相等
-dp[i][j]=min{
-    dp[i-1][j],
-    dp[i][j-1],
-    dp[i-1][j-1]
-}+1
-
-dp[i][j]=INT_MAX
-dp[0][0]=0
-dp[i][0]=i
-dp[0][j]=j
-*/
-#include <algorithm>
-#include <climits>
-#include <cstdint>
-#include <iostream>
-#include <istream>
-#include <string>
-#include <vector>
-using ll = int64_t;
-
-int main(){
-    std::iostream::sync_with_stdio(false);
-    std::cin.tie(nullptr);
-    std::string s1,s2;
-    std::cin>>s1>>s2;
-    ll n=s1.size(),m=s2.size();
-    s1=' '+s1;
-    s2=' '+s2;
-    std::vector<std::vector<ll>> dp(s1.size(),std::vector<ll>(s2.size(),INT_MAX));
-    for(ll i=0;i<dp.size();i++)dp[i][0]=i;
-    for(ll j=0;j<dp[0].size();j++)dp[0][j]=j;
-    for(ll i=1;i<=n;i++){
-        for(ll j=1;j<=m;j++){
-            if(s1[i]==s2[j]){
-                dp[i][j]=dp[i-1][j-1];
-            }else{
-                dp[i][j]=std::min({
-                    dp[i-1][j],
-                    dp[i][j-1],
-                    dp[i-1][j-1]
-                })+1;
-            }
-        }
-    }
-    std::cout<<dp[n][m]<<"\n";
-}

src/8/26/P2782.cpp

@@ -41,7 +41,6 @@ int main(){
         } else {
             *it = x;
         }
-        pa(f);
     }
     std::cout << f.size() << std::endl;
     return 0;

src/8/26/P5124.cpp

@@ -1,27 +1,28 @@
 #include <algorithm>
-#include <cstdint>
 #include <iostream>
 #include <istream>
-#include <ostream>
+#include <numeric>
 #include <vector>
+
 using ll = int64_t;
 
+#define pv(v)do{std::cout<<#v<<": "<<(v)<<"\n";}while(0)
+
 int main(){
     std::iostream::sync_with_stdio(false);
     std::cin.tie(nullptr);
+
     ll n,k;
     std::cin>>n>>k;
-    std::vector<ll> v(n+1),dp(n+1);
+    std::vector<ll> v(n);
-    for(ll i=1;i<=n;i++){
+    for(ll&i:v){
-        std::cin>>v[i];
+        std::cin>>i;
     }
-    for(ll i=1;i<=n;i++){
+    std::sort(v.begin(),v.end());
-        ll max=0;
+    ll teams=n/k,last=n%k;
-        for(ll len=1;len<=k&&len<=i;len++){
+    ll teamssum = std::accumulate(v.rbegin(),v.rbegin()+teams,0);
-            max=std::max(max,v[i-len+1]);
+    // pv(teams);
-            dp[i]=std::max(dp[i],max*len+dp[i-len]);
+    // pv(teamssum);
-        }
+    // pv(last);
-    }
+    std::cout<<(last * (*(v.rbegin()+teams)) + teamssum*k)<<"\n";
-    std::cout<<dp[n]<<std::endl;
-    _Exit(0);
 }

src/8/26/P5424.cpp

@@ -1,68 +0,0 @@
-/*
-
-dp[i][j]
-前i组蛇，用j个网的最小浪费空间
-
-dp[i][j]=min(dp[i][j],dp[m][j-1]+cnt[m+1][i])
-
-dp[i][j]=1e9
-dp[0][0]=0
-
-*/
-
-#include <algorithm>
-#include <cstdint>
-#include <cstdio>
-#include <iostream>
-#include <istream>
-#include <vector>
-
-using ll = int64_t;
-
-#define p2v(dp)do{\
-    for(ll i=0;i<dp.size();i++){\
-        for(ll j=0;j<dp[i].size();j++){\
-            printf("[%lld][%lld]=%lld\n",i,j,dp[i][j]);\
-        }\
-    }\
-}while(0)
-
-int main(){
-    std::iostream::sync_with_stdio(false);
-    std::cin.tie(nullptr);
-
-    ll n,m;
-    std::cin>>n>>m;
-    m++;
-    std::vector<ll> v(n+1);
-    std::vector<std::vector<ll>> cnt(n+1,std::vector<ll>(n+1));//cnt[i][j]=i-j用一个网浪费的空间
-    std::vector<std::vector<ll>> dp(n+1,std::vector<ll>(m+1, 1e9));
-    dp[0][0]=0;
-
-    for(ll i=1;i<=n;i++){
-        std::cin>>v[i];
-    }
-    for(ll i=1;i<=n;i++){
-        for(ll j=i;j<=n;j++){
-            ll nmax = *std::max_element(v.begin()+i,v.begin()+j+1);
-            ll ncnt=0;
-            for(ll m=i;m<=j;m++){
-                ncnt+=nmax-v[m];
-            }
-            cnt[i][j]=ncnt;
-        }
-    }
-    // printf("cnt\n");
-    // p2v(cnt);
-    for(ll i=1;i<=n;i++){
-        dp[i][1]=cnt[1][i];
-        for(ll j=2;j<=m&&j<=i;j++){
-            for(ll k=j-1;k<i;k++){
-                dp[i][j]=std::min(dp[i][j],dp[k][j-1]+cnt[k+1][i]);
-            }
-        }
-    }
-    // printf("dp\n");
-    // p2v(dp);
-    std::cout<<dp[n][m]<<"\n";
-}

src/8/26/P7414.cpp

@@ -1,6 +0,0 @@
-/*
-
-dp[i][j]=从第i个到第j个涂成指定颜色需要的次数
-
-
-*/