[数位DP] BZOJ 2713 [Violet 2]愚蠢的副官 BZOJ 1183 [Croatian2008]Umnozak

官方题解

The first observation we can make is that, for all positive integers x x, the digit-product p(x)p(x) is always less than or equal to x x. From x∗p(x)≤B≤1018x*p(x) ≤ B ≤ 10^{18} we can deduce that p(x)≤B≤109 p(x) ≤ B ≤ 10^9
Because p(x) p(x) is a product of digits, its prime factors can only be 2 2, 33, 5 5 and 77. The number of different integers with only 2 2, 33, 5 5 and 77 in their prime factorization that are less than or equal to 109 10^9 is quite small (exactly 5194 5194).

然后就很显然了
枚举出 p(x) p(x) 然后找小于等于 n/p(x) n/p(x)的满足条件的 x <script type="math/tex" id="MathJax-Element-2406">x</script>
简单的数位DP 我不顾及复杂度地乱写一通竟然过了
~~为什么std跑的那么快嘤嘤嘤~~

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;const int fact[]={2,3,5,7};
const int C[11][4]={  {0,0,0,0}, {0,0,0,0}, {1,0,0,0}, {0,1,0,0}, {2,0,0,0}, {0,0,1,0}, {1,1,0,0}, {0,0,0,1}, {3,0,0,0}, {0,2,0,0}
};
int f[25][45][35][25][25];inline ll count(ll n,int a,int b,int c,int d){if (!n) return 0;int w[20]={0},m=0; n++;for (ll t=n;t;w[++m]=t%10,t/=10); reverse(w+1,w+m+1);for (int t=1;t<=m;t++)for (int i=0;i<=a;i++)for (int j=0;j<=b;j++)for (int k=0;k<=c;k++)for (int l=0;l<=d;l++) f[t][i][j][k][l]=0;int p[4]={0}; int flag=0;for (int t=1;t<=m;t++){for (int r=1;r<=9;r++)for (int i=0;i+C[r][0]<=a;i++)for (int j=0;j+C[r][1]<=b;j++)for (int k=0;k+C[r][2]<=c;k++)for (int l=0;l+C[r][3]<=d;l++)f[t][i+C[r][0]][j+C[r][1]][k+C[r][2]][l+C[r][3]]+=f[t-1][i][j][k][l];if (!flag)for (int r=1;r<w[t];r++)f[t][p[0]+C[r][0]][p[1]+C[r][1]][p[2]+C[r][2]][p[3]+C[r][3]]++;if (t!=1){for (int r=1;r<=9;r++)f[t][C[r][0]][C[r][1]][C[r][2]][C[r][3]]++;}if (!w[t])flag=1;elsefor (int i=0;i<4;i++)p[i]+=C[w[t]][i];}return f[m][a][b][c][d];
}inline ll Solve(ll n){if (!n) return 0;ll a,b,c,d; int i,j,k,l; ll B=sqrt(n);ll ret=0;for (i=0,a=1;a<=B;i++,a*=fact[0])for (j=0,b=a;b<=B;j++,b*=fact[1])for (k=0,c=b;c<=B;k++,c*=fact[2])for (l=0,d=c;d<=B;l++,d*=fact[3])ret+=count(n/d,i,j,k,l);return ret;
}int main(){ll L,R;freopen("t.in","r",stdin);freopen("t.out","w",stdout);scanf("%lld%lld",&L,&R);printf("%lld\n",Solve(R)-Solve(L-1));return 0;
}

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