refactor(P3102): 重构字符串处理逻辑为动态规划实现

将原有的暴力匹配算法重构为基于动态规划的解决方案，提高计算效率
添加了调试输出和模数运算支持

src/11/13/P3102.cpp

@@ -1,34 +1,57 @@
 #include <algorithm>
 #include <cstdint>
+#include <cstdio>
 #include <iostream>
 #include <istream>
 #include <string>
+#include <vector>
 using ll = int64_t;
-
+#define printf
+const ll p=2014;
 std::string s;
-ll ans;
+ll n;
+std::vector<std::vector<ll>> dp;
 
-int main(){
-    std::cin.tie(nullptr);
-    std::iostream::sync_with_stdio(false);
-
-    std::cin>>s;
-    for(ll i=1;i<s.size();i++){
-        ll l1=i,l2=s.size()-i;
-        for(ll j=0;j<std::min(l1,l2);j++){
-            if(s[j]!=s[i+j]){
-                goto nxt1;
-            }
+static inline ll dfs(const ll l,const ll r){
+    struct S{
+        const ll &l,&r,&dp;
+        S(const ll&l,const ll&r,const ll&dp):l(l),r(r),dp(dp){
+            printf("dfs l=%lld, r=%lld, START\n",l,r);
         }
-        ans++;
+        ~S(){
+            printf("dfs l=%lld, r=%lld, ret=%lld\n",l,r,dp);
+        }
+    }raii(l,r,dp[l][r]);
+    const ll nlen=r-l+1;
+    if(l>r)return 0;
+    if(dp[l][r])return dp[l][r];
+    dp[l][r]=1;
+    if(nlen<=2)return dp[l][r]=1;
+    for(ll i=l+1;i<=r;i++){
+        if(i-l==r-i+1)continue;
+        const ll clen=std::min(i-l,r-i+1);
+        if(i-l==1&&r-i+1==1)continue;
+        for(ll j=0;j<clen;j++){
+            if(s[l+j]!=s[i+j])goto nxt1;
+        }
+        dp[l][r]=((i-l>r-i+1?dfs(l, i-1):dfs(i, r))+dp[l][r])%p;
         nxt1:;
-        for(ll j=0;j<std::min(l1,l2);j++){
-            if(s[i-1-j]!=s[s.size()-1-j]){
-                goto nxt2;
-            }
+        for(ll j=0;j<clen;j++){
+            if(s[i-1-j]!=s[r-j])goto nxt2;
         }
-        ans++;
+        dp[l][r]=((i-l>r-i+1?dfs(l, i-1):dfs(i, r))+dp[l][r])%p;
         nxt2:;
     }
-    std::cout<<ans*2<<"\n";
+    return dp[l][r];
+}
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    std::cin>>s;
+    n=s.size();
+    s=' '+s;
+    dp.resize(n+1,std::vector<ll>(n+1));
+    std::cout<<(dfs(1,n)-1+p)%p<<"\n";
 }