feat: 添加动态规划解法用于解决特定问题

实现了一个基于动态规划的算法来解决给定问题，包括处理输入数据、计算最优解和输出结果。主要逻辑集中在solve函数中，通过遍历和比较来找到最优解。

src/11/17/P14508.cpp

@@ -0,0 +1,100 @@
+#include <algorithm>
+#include <cstdint>
+#include <cstdio>
+#include <iostream>
+#include <istream>
+#include <map>
+#include <vector>
+using ll = int64_t;
+#define sl static inline
+#define printf
+const ll inf=1e9+7;
+std::vector<ll> a,b,dp,dpf;
+std::map<ll, ll> rstep;
+ll n,m;
+struct Mu{
+    ll num,idx;
+    inline bool operator<(const Mu&o)const{
+        return num<o.num;
+    }
+};
+std::vector<Mu> cminuse;
+
+sl ll minuse(ll val){
+    if(cminuse[val].idx!=inf){
+        return cminuse[val].idx;
+    }
+    for(ll i=1;i<=m;i++){
+        const ll num = (val+a[i]-1)/a[i];
+        cminuse[val]=std::min(cminuse[val],{num*b[i],i});
+    }
+    return cminuse[val].idx;
+}
+
+sl void solve(){
+    ll ans=0;
+    std::cin>>n>>m;
+    a.clear(),b.clear(),dp.clear(),dpf.clear(),rstep.clear(),cminuse.clear();
+    a.resize(m+1),b.resize(m+1),dp.resize(n+1,inf),dpf.resize(n+1,inf),cminuse.resize(n+1,{inf,inf});
+    dp[0]=0;
+    for(ll i=1;i<=m;i++){
+        std::cin>>a[i];
+    }
+    for(ll i=1;i<=m;i++){
+        std::cin>>b[i];
+    }
+    for(ll i=1;i<=m;i++){
+        for(ll j=a[i];j<=n;j++){
+            const ll ndp=dp[j-a[i]]+b[i];
+            // printf("i=%lld, j=%lld, ndp=%lld\n",i,j,ndp);
+            if(ndp<dp[j]){
+                // printf("got new min %lld\n",ndp);
+                dp[j]=ndp;
+                dpf[j]=j-a[i];
+            }
+        }
+    }
+    for(ll i=1;i<=m;i++){
+        for(ll j=n-a[i];j>=1;j--){
+            dp[j]=std::min(dp[j],dp[j+a[i]]+b[i]);
+        }
+    }
+    printf("dp[]: ");
+    for(ll i=1;i<=n;i++){
+        printf("%lld ",dp[i]);
+    }
+    printf("\n");
+    printf("dpf[]: ");
+    for(ll i=1;i<=n;i++){
+        printf("%lld ",dpf[i]);
+    }
+    printf("\n");
+    printf("minuse(%lld)=%lld\n",n,minuse(n));
+    ll nuse=0;
+    for(ll i=1;i<=n;){
+        const ll use=minuse(n-i),k=i+a[use];
+        nuse+=b[use];
+        for(ll j=i+1;j<=k&&j<=n;j++){
+            if(dp[k-j]==inf){
+                std::cout<<"-1\n";
+                return;
+            }else{
+                ans=std::max(ans,dp[k-j]+nuse);
+            }
+        }
+        i=k;
+    }
+    std::cout<<ans<<"\n";
+}
+
+int main(){
+    #ifdef printf
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+    #endif
+    ll c,t;
+    std::cin>>c>>t;
+    while(t--){
+        solve();
+    }
+}