【动态规划】【记忆化搜索】CODEVS 最小和 CodeVS原创

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 int save[9][9][9],m,n,wei,ans=2147483647,len;
 6 const int Base[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};
 7 int Get_Part(const int &l,const int &r) {return n%Base[wei-l+1]/Base[wei-r];}
 8 int f(int sta,int end,int now)
 9 {
10     if(save[sta][end][now]!=-1) return save[sta][end][now];
11     int res=2147483647;
12     if(now==2) res=min(res,Get_Part(sta,end)+f(end+1,wei,now-1));
13     else
14       for(int i=end+1;i<=wei-(now-2);i++)
15         res=min(res,Get_Part(sta,end)+f(end+1,i,now-1));
16     return save[sta][end][now]=res;
17 }
18 int main()
19 {
20     scanf("%d%d",&n,&m);
21     if(m==1)
22       {
23           printf("%d\n",n);
24           return 0;
25       } int t=n;
26     while(t) {wei++; t/=10;}
27     memset(save,-1,sizeof(save));
28     for(int i=1;i<=wei-m+1;i++)
29       save[wei-i+1][wei][1]=Get_Part(wei-i+1,wei);
30     for(int i=1;i<=wei-m+1;i++)
31       ans=min(ans,f(1,i,m));
32     printf("%d\n",ans);
33     return 0;
34 }