CodeForces 258B Little Elephant and Elections ：于1-m中找出七个数，使六个数里面的4和7个数比第七个数严格小：数位dp+dfs
预处理sum数组，sum[i]表示1-m中有i个4或7的数有多少个，这个数位dp很好写
然后就是枚举第七个数含有的4,7数目，dfs剩下的六个数=
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 using namespace std; 5 #define LL long long 6 #define MOD 1000000007 7 LL num[15],dp[15][15][15],sum[15]; 8 LL dfs(LL pos,LL pre,LL sum,LL limit) 9 { 10 LL ans=0; 11 if (pos==0) return pre==sum; 12 if (!limit&&dp[pos][pre][sum]!=-1) 13 return dp[pos][pre][sum]; 14 LL tmp=limit?num[pos]:9,i; 15 for (i=0;i<=tmp;i++) 16 ans+=dfs(pos-1,pre+(i==4||i==7),sum,limit&&i==tmp); 17 if (!limit) dp[pos][pre][sum]=ans; 18 return ans; 19 } 20 LL solve(LL now,LL res) 21 { 22 LL ans=0,i; 23 if (now==7) return 1; 24 for (i=0;i<=res;i++) 25 if (sum[i]>0) 26 { 27 sum[i]--; 28 ans=(ans+(sum[i]+1)*solve(now+1,res-i))%MOD; 29 sum[i]++; 30 } 31 return ans; 32 } 33 int main() 34 { 35 LL m,cnt=0,ans=0,i; 36 scanf("%I64d",&m); 37 memset(dp,-1,sizeof(dp)); 38 while (m){ 39 num[++cnt]=m%10; 40 m/=10; 41 } 42 for (i=0;i<=cnt;i++) sum[i]=dfs(cnt,0,i,1); sum[0]--; 43 for (i=1;i<=cnt;i++) 44 ans=(ans+sum[i]*solve(1,i-1))%MOD; 45 printf("%I64d\n",ans); 46 return 0; 47 }
