[CodeForces908G]New Year and Original Order

题意:定义函数$S(x)$是把$x$各数位排序后得到的数,求$\sum_{i=1}^nS(i)$

$S(2018)=128,S(998244353)=233445899$

$n\le10^{700}$

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看着$n$这么大显然是数位$DP$了

扒一下官方题解

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This is a digit $dp$ problem. Let’s try to solve the subproblem “How many ways can the$i-th$ digit be at least $j$?”. Let’s fix $j$, and solve this with $dp$. We have a dp state $dp[a][b][c]$=number of ways given we’ve considered the first a digits of $X$, we need $b$ more occurrences of digits at least $j$, and $c$ is a boolean saying whether or not we are strictly less than $X$ or not yet.

For a fixed digit, we can compute this dp table in $O(n^2)$ time, and then compute the answers to our subproblem for each $i$ ($i.e.^1$ by varying $b$ in our table).

1.$i.e.$ $\Rightarrow$in other words/this is

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考虑到原问题很难做,我们转换一下问题

举个例子:假设第$i$位填$3$,那么他的贡献的$3*10^i$,考虑把这个贡献拆开

$3\ge3,3\ge2,3\ge1$我们在这一位填$1,2,3$时都记上$10^i$的贡献

那么我们就只要记这一位填大于等于某个数的方案数就好了

考虑按照套路设$dp[i][j][k][l]$表示前$i$位有$j$位的数字大小大于等于$k$,是否严格小于$n$的方案数

枚举第$i+1$位填$p$

$f[i+1][j+(p>=k)][k][l|(p<a_{i+1})]=\sum f[i][j][k][l]$

然后实际上假设前$n$位我们$j$位数字大于等于$k$的方案数是$sum=f[n][j][k][0]+f[n][j][k][1]$

这个对答案的贡献的是$sum*\underbrace{111\ldots11}_{j\text{个}1}$

 #include<bits/stdc++.h>
#define fp(i,a,b) for(register int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(register int i=a,I=b-1;i>I;--i)
#define go(u) for(register int i=fi[u],v=e[i].to;i;v=e[i=e[i].nx].to)
#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
using namespace std;
const int N=705,P=1e9+7;
int n,ans,a[N],f[N][N][10][2];char s[N];
inline void add(int&a,int b){a+=b,a>=P?a-=P:0;}
int main(){
    #ifndef ONLINE_JUDGE
        file("s");
    #endif
    scanf("%s",s+1);n=strlen(s+1);
    fp(i,1,n)a[i]=s[i]-48;
    fp(i,0,9)f[0][0][i][0]=1;
    fp(i,0,n-1)fp(j,0,i)fp(k,1,9)fp(l,0,1)fp(p,0,(l?9:a[i+1]))
         add(f[i+1][j+(p>=k)][k][l|(p<a[i+1])],f[i][j][k][l]);
    fp(k,1,9){
        int tp=1;
        fp(i,1,n)add(ans,1ll*tp*(f[n][i][k][0]+f[n][i][k][1])%P),tp=(10ll*tp+1)%P;
    }printf("%d",ans);
return 0;
}

发现这个转移是可以滚动的,而且每一个还可以分开考虑贡献

所以泥萌可以自己再卡一卡常数