数位dp

1.[hdu3709]Balanced Number

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<string>
 5 #include<cstdlib>
 6 #include<algorithm>
 7 #include<ctime>
 8 #include<cmath>
 9 #include<queue>
10 #include<map>
11 #include<set>
12 using namespace std;
13 #define FILE "dealing"
14 #define LL long long
15 #define up(i,j,n) for(LL i=(j);(i)<=(n);i++)
16 #define pii pair< LL , LL >
17 #define abs(x) ((x)<0?-(x):(x))
18 #define max(x,y) ((x)>(y)?(x):(y))
19 #define min(x,y) ((x)<(y)?(x):(y))
20 namespace IO{
21     char buf[1<<15],*fs,*ft;
22     LL gc(){return fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft),fs==ft?-1:*fs++;}
23     LL read(){
24         LL x=0,ch=gc(),f=0;
25         while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=1;ch=gc();}
26         while(ch>=‘0‘&&ch<=‘9‘){x=(x<<1)+(x<<3)+ch-‘0‘;ch=gc();}
27         return f?-x:x;
28     }
29 }using namespace IO;
30 bool chkmin(LL &a,LL b){return a>b?a=b,true:false;}
31 bool chkmax(LL &a,LL b){return a<b?a=b,true:false;}
32 const LL maxn(1000500),inf(100000000);
33 LL f[20][20][2500],now[20];
34 LL dfs(LL pos,LL pivot,LL ans,LL limit){
35     if(pos<1)return ans == 0;
36     if(ans<0)return 0;
37     if(!limit&&f[pos][pivot][ans]!=-1)return f[pos][pivot][ans];
38     LL len=limit?now[pos]:9;
39     LL ret=0;
40     up(i,0,len)ret+=dfs(pos-1,pivot,ans+(pos-pivot)*i,limit&&i==len);
41     if(!limit)f[pos][pivot][ans]=ret;
42     return ret;
43 }
44 LL slove(LL x){
45     now[0]=0;
46     while(x){now[++now[0]]=x%10;x/=10;}
47     LL ret=0;
48     up(i,1,now[0])ret+=dfs(now[0],i,0,1);
49     return ret-now[0]+1;
50 }
51 int main(){
52     //freopen(FILE".in","r",stdin);
53     //freopen(FILE".out","w",stdout);
54     LL A,B;
55     LL T;
56     T=read();
57     memset(f,-1,sizeof(f));
58     while(T--){
59         A=read();B=read();
60         if(A==0)cout<<slove(B)<<endl;
61         else cout<<slove(B)-slove(A-1)<<endl;
62     }
63     return 0;
64 }

记忆化搜索

2，[hdu3555]只要49

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<string>
 5 #include<cstdlib>
 6 #include<algorithm>
 7 #include<ctime>
 8 #include<cmath>
 9 #include<queue>
10 #include<map>
11 #include<set>
12 using namespace std;
13 #define FILE "dealing"
14 #define LL long long
15 #define up(i,j,n) for(LL i=(j);(i)<=(n);i++)
16 #define pii pair< LL , LL >
17 #define abs(x) ((x)<0?-(x):(x))
18 #define max(x,y) ((x)>(y)?(x):(y))
19 #define min(x,y) ((x)<(y)?(x):(y))
20 namespace IO{
21     char buf[1<<15],*fs,*ft;
22     LL gc(){return fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft),fs==ft?-1:*fs++;}
23     LL read(){
24         LL x=0,ch=gc(),f=0;
25         while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=1;ch=gc();}
26         while(ch>=‘0‘&&ch<=‘9‘){x=(x<<1)+(x<<3)+ch-‘0‘;ch=gc();}
27         return f?-x:x;
28     }
29 }using namespace IO;
30 bool chkmin(LL &a,LL b){return a>b?a=b,true:false;}
31 bool chkmax(LL &a,LL b){return a<b?a=b,true:false;}
32 const LL maxn(1000500),inf(100000000);
33 LL f[50][10];
34 void init(){
35     up(i,0,9)f[1][i]=1;
36     up(i,2,20)up(j,0,9){
37         up(k,0,9){
38             if(j==4&&k==9)continue;
39             f[i][j]+=f[i-1][k];
40         }
41     }
42 }
43 LL slove(LL x){
44     LL now[21],ans=0,flag=0,y=x;
45     now[0]=0;
46     while(x){now[++now[0]]=x%10;x/=10;}
47     up(i,1,now[0]-1)up(j,1,9)ans+=f[i][j];
48     up(i,1,now[now[0]]-1)ans+=f[now[0]][i];
49     LL pre=now[now[0]];
50     for(LL i=now[0]-1;i>=1;i--){
51         if(now[i]==9&&now[i+1]==4)flag=1;
52         up(j,0,now[i]-1){
53             if(now[i+1]==4&&j==9)continue;
54             ans+=f[i][j];
55         }
56         if((now[i]==9&&now[i+1]==4))break;
57         pre=now[i];
58     }
59     if(!flag&&y)ans++;
60     return ans;
61 }
62 int main(){
63     //freopen(FILE".in","r",stdin);
64     //freopen(FILE".out","w",stdout);
65     init();LL A,B;
66     LL T;
67     T=read();
68     while(T--){
69         A=read();
70         cout<<A-slove(A)<<endl;
71     }
72     return 0;
73 }

递推

3.[hdu2089]不要62

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<string>
 5 #include<cstdlib>
 6 #include<algorithm>
 7 #include<ctime>
 8 #include<cmath>
 9 #include<queue>
10 #include<map>
11 #include<set>
12 using namespace std;
13 #define FILE "dealing"
14 #define LL long long
15 #define up(i,j,n) for(int i=(j);(i)<=(n);i++)
16 #define pii pair< int , int >
17 #define abs(x) ((x)<0?-(x):(x))
18 #define max(x,y) ((x)>(y)?(x):(y))
19 #define min(x,y) ((x)<(y)?(x):(y))
20 namespace IO{
21     char buf[1<<15],*fs,*ft;
22     int gc(){return fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft),fs==ft?-1:*fs++;}
23     int read(){
24         int x=0,ch=gc(),f=0;
25         while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=1;ch=gc();}
26         while(ch>=‘0‘&&ch<=‘9‘){x=(x<<1)+(x<<3)+ch-‘0‘;ch=gc();}
27         return f?-x:x;
28     }
29 }using namespace IO;
30 bool chkmin(int &a,int b){return a>b?a=b,true:false;}
31 bool chkmax(int &a,int b){return a<b?a=b,true:false;}
32 const int maxn(1000500),inf(100000000);
33 int f[50][10];
34 void init(){
35     up(i,0,9)f[1][i]=1;f[1][4]=0;
36     up(i,2,10)up(j,0,9){
37         if(j==4)continue;
38         up(k,0,9){
39             if(j==6&&k==2)continue;
40             f[i][j]+=f[i-1][k];
41         }
42     }
43 }
44 int slove(int x){
45     int now[11],ans=0,flag=0;
46     now[0]=0;
47     if(x==0)return 0;
48     while(x){now[++now[0]]=x%10;x/=10;}
49     up(i,1,now[0]-1)up(j,1,9)ans+=f[i][j];
50     up(i,1,now[now[0]]-1)ans+=f[now[0]][i];
51     int pre=now[now[0]];
52     if(pre==4)return ans;
53     for(int i=now[0]-1;i>=1;i--){
54         if(now[i]==4||(now[i]==2&&now[i+1]==6))flag=1;
55         up(j,0,now[i]-1){
56             if(now[i+1]==6&&j==2)continue;
57             ans+=f[i][j];
58         }
59         if((now[i]==2&&now[i+1]==6)||now[i]==4)break;
60         pre=now[i];
61     }
62     if(!flag)ans++;
63     return ans;
64 }
65 int main(){
66     //freopen(FILE".in","r",stdin);
67     //freopen(FILE".out","w",stdout);
68     init();int A,B;
69     while(1){
70         A=read(),B=read();
71         if(A==0&&B==0)return 0;
72         printf("%d\n",slove(B)-slove(A-1));
73     }
74     return 0;
75 }

递推

4.[haoi2010]计数

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<ctime>
 7 #include<iomanip>
 8 #include<queue>
 9 #include<map>
10 #include<set>
11 #include<algorithm>
12 using namespace std;
13 #define FILE "dealing"
14 #define LL long long
15 #define up(i,j,n) for(int i=(j);i<=(n);i++)
16 void chkmin(int &a,int b){a>b?a=b:0;}
17 void chkmax(int &a,int b){a<b?a=b:0;}
18 namespace IO{
19     char buf[1<<15],*fs,*ft;
20     int gc(){return (fs==ft&&(ft=(fs=buf)+fread(buf,1<<15,1,stdin),fs==ft))?-1:*fs++;}
21     int read(){
22         int x=0,f=0,ch=gc();
23         while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=1;ch=gc();}
24         while(ch>=‘0‘&&ch<=‘9‘){x=(x<<1)+(x<<3)+ch-‘0‘;ch=gc();}
25         return f?-x:x;
26     }
27 }using namespace IO;
28 const int maxn=52;
29 char s[maxn];
30 int a[10],len;
31 LL fac[30],c[maxn][maxn];
32 int main(){
33     freopen(FILE".in","r",stdin);
34     freopen(FILE".out","w",stdout);
35     c[0][0]=1;
36     for(int i=1;i<=50;i++){
37         c[i][0]=1;
38         for(int j=1;j<=50;j++)c[i][j]=c[i-1][j-1]+c[i-1][j];
39     }
40     scanf("%s",s+1);
41     len=strlen(s+1);
42     up(i,1,len)a[s[i]-‘0‘]++;
43     LL ans=0;
44     for(int i=1;i<=len;i++){
45         for(int j=0;j<s[i]-‘0‘;j++){
46             if(!a[j])continue;
47             a[j]--;
48             int now=len-i;
49             LL sum=1;
50             for(int k=0;k<=9;k++)
51                 sum*=c[now][a[k]],now-=a[k];
52             ans+=sum;
53             a[j]++;
54         }
55         a[s[i]-‘0‘]--;
56     }
57     cout<<ans<<endl;
58     return 0;
59 }

递推

通过这几道题的练习，对数位dp有了些理解；

数位dp的核心是按位确定，这些题无论约束条件如何，但本质不变；

有的约束条件用递推可以解决，于是可以预处理，有的约束条件比较复杂，需要记忆化；

数位dp也体现了dp的本质，每个数位其实就是一个阶段，每次搜索即将发散的时候，下面都有一个状态的承接，不至于时间爆炸；