Good Bye 2017 G. New Year and Original Order

G. New Year and Original Order

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let S(n) denote the number that represents the digits of n in sorted order. For example, S(1) = 1, S(5) = 5, S(50394) = 3459, S(353535) = 333555.

Given a number X, compute modulo 10⁹ + 7.

Input

The first line of input will contain the integer X (1 ≤ X ≤ 10⁷⁰⁰).

Output

Print a single integer, the answer to the question.

Examples

Input

21

Output

195

Input

345342

Output

390548434

Note

The first few values of S are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2, 12. The sum of these values is 195.

 

tls原话：最简单的数位dp，出题人成功骗过验题人放在了G。

Oh...uuu...原来是简单的数位dp...等等...tls的简单，莫不是.....

#include<bits/stdc++.h>
#define LL long long
using namespace std;
char s[705];
int dp[705][705][2],res;
const int mod=1e9+7;
int main()
{
    cin>>s;
    int len=strlen(s);
    int ll,kk,i,j,k,l,m;
    for(i=1;i<=9;i++)
    {
        dp[0][0][1]=1;
        for(j=0;j<len;j++)
            for(k=0;k<=j;k++)
                for(l=0;l<=1;l++)
                    for(m=0;m<=9;m++)
                    {
                        if(m<s[j]-'0')  ll=0;
                        else if(m==s[j]-'0')    ll=l;
                        else if(l)  continue;
                        else    ll=0;
                        if(m>=i)  kk=k+1;
                        else kk=k;
                        dp[j+1][kk][ll]=(dp[j+1][kk][ll]+dp[j][k][l])%mod;
                    }
        for(j=1,k=1;j<=len;j++)
        {
            res=(res+(LL)k*(dp[len][j][0]+dp[len][j][1]))%mod;
            k=(k*10ll+1)%mod;
        }
        for(j=0;j<=len;j++)
            for(k=0;k<=j;k++)
                for(l=0;l<=1;l++)
                    dp[j][k][l]=0;
    }
    cout<<res<<endl;
    return 0;
}

九月巨巨的滚动数组：

#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7,maxn=710;
#define LL long long
LL sum[maxn][2],dp[maxn][2];
string s;
inline void Update(LL &x,LL y)
{
   x=(x+y)%mod;
}
void Clear()
{
    memset(dp,0,sizeof(dp));
    memset(sum,0,sizeof(sum));
    dp[0][0]=1;
}
int main()
{
    cin>>s;
    int len=s.length();
    LL ans=0;
    for(int l=1;l<=9;l++)
    {
        Clear();
        for(int i=0;i<len;i++)
            for(int j=0;j<2;j++)
                for(int k=0;k<=9;k++)
                {
                    if(!j&&k>s[i]-'0') continue;
                    Update(dp[i+1][j|k<s[i]-'0'],dp[i][j]);  ///j和k<s[i]  任意一个是1就为1
                    if(k<l) Update(sum[i+1][j|k<s[i]-'0'],sum[i][j]);
                    else Update(sum[i+1][j|k<s[i]-'0'],(sum[i][j]*10+dp[i][j])%mod);
                }
        Update(ans,sum[len][0]);
        Update(ans,sum[len][1]);
    }
    cout<<ans<<endl;
    return 0;
}