20140712 classic 数位DP

 1 #include <cstdio>
 2 #include <cstring>
 3 using namespace std;
 4 #define mod 1000000007
 5 typedef long long ll;
 6 
 7 ll dp[60][600][2];
 8 ll num[60][600][2];
 9 int x,y;
10 
11 ll DP(char ch[60]) {
12     memset(dp,0,sizeof(dp));
13     memset(num,0,sizeof(num));
14     int len=strlen(ch);
15     int a[60];
16     int sum[60];
17     ll ans=0;
18     sum[0]=0;
19     for (int i=0;i<len;i++) {
20         a[i+1]=(int)(ch[i])-48;
21         sum[i+1]=sum[i]+a[i+1];
22     }
23     for (int j=0;j<a[1];j++) {
24         dp[1][j][0]=j; num[1][j][0]=1;
25     }
26     dp[1][a[1]][1]=a[1]; num[1][a[1]][1]=1;
27     for (int i=1;i<len;i++) {
28         for (int j=0;j<600;j++) {
29             if (j!=0 && dp[i][j][0]==0) continue;
30             for (int k=0;k<10;k++) {
31                 num[i+1][j+k][0]=(num[i+1][j+k][0]+num[i][j][0])%mod;
32                 dp[i+1][j+k][0]=(dp[i+1][j+k][0]+dp[i][j][0]*10+k*num[i][j][0])%mod;
33             }
34         }
35         if (sum[i]!=0 && dp[i][sum[i]][1]==0) continue;
36         for (int k=0;k<a[i+1];k++) {
37             num[i+1][sum[i]+k][0]=(num[i+1][sum[i]+k][0]+num[i][sum[i]][1])%mod;
38             dp[i+1][sum[i]+k][0]=(dp[i+1][sum[i]+k][0]+dp[i][sum[i]][1]*10+k*num[i][sum[i]][1])%mod;
39         }
40         num[i+1][sum[i+1]][1]=(num[i][sum[i]][1]+num[i+1][sum[i+1]][1])%mod;
41         dp[i+1][sum[i+1]][1]=(dp[i+1][sum[i+1]][1]+dp[i][sum[i]][1]*10+a[i+1]*num[i][sum[i]][1])%mod;
42     }
43     for (int j=x;j<=y;j++) ans=(ans+dp[len][j][0]+dp[len][j][1])%mod;
44     return ans;
45 }
46 
47 char l[60],r[60];
48 
49 int main() {
50     scanf("%s",&l);
51     scanf("%s",&r);
52     scanf("%d%d",&x,&y);
53     int len=strlen(l);
54     len--;
55     while (1) {
56         if (l[len]==‘0‘) l[len]=‘9‘;
57         else {
58             l[len]=(char(int(l[len])-1));
59             break;
60         }
61         len--;
62         if (len==0) {
63             for (int i=strlen(l)-1;i>0;i--) {
64                 l[i-1]=l[i];
65             }
66             l[strlen(l)-1]=‘\n‘;
67             break;
68         }
69     }
70     ll R=DP(r);
71     ll L=DP(l);
72     while (R-L<0) R+=mod;
73     printf("%lld",(R-L)%mod);
74 }