[swustoj 917] K-lucky-number

K-lucky-number(0917)

问题描述

K-lucky-number is defined as add up the number of each bit is a multiple of K.for example, 24 is a 3-lucky-number,because 2+4=6, 6 is a multiple of 3.Now, there is a closed interval from L to R, please output the sum of squares of the K-lucky-number in this interval.

输入

The first line of input is the number of test cases T.

For each test case have one line contains three integer L, R, K(0<L<=R<10^9, 1<k<30).

输出

For each test case output the answer the sum of squares of the K-lucky-number in this interval and mod 1000000007.

样例输入

2
1 10 6
100 1000 7

样例输出

36
46057247

有点6、= =

#include <iostream>

#include <cstring>

#include <algorithm>

#include <cstdio>

#include <cmath>

using namespace std;

#define ll long long

#define MOD 1000000007

#define INF 0x3f3f3f3f

#define N 50

 

struct Node{

    ll cnt,sum,sqsum; //个数,和,平方和

};

 

int l,r,k;

int bit[N];

ll p10[N];

Node dp[N][N];

 

void init()

{

    p10[0]=1;

    for(int i=1;i<=20;i++) p10[i]=p10[i-1]*10%MOD;

}

Node dfs(int pos,int mod,bool limit)

{

    Node ans;

    ans.cnt=ans.sum=ans.sqsum=0;

    if(pos==-1){

        if(!mod) ans.cnt=1;

        else ans.cnt=0;

        return ans;

    }

    if(!limit && dp[pos][mod].sum!=-1) return dp[pos][mod];

    int end=limit?bit[pos]:9;

    for(int i=0;i<=end;i++){

        Node tmp=dfs(pos-1,(mod+i)%k,(i==end)&&limit);

        ans.cnt=(ans.cnt+tmp.cnt)%MOD;

        ans.sum=(ans.sum + tmp.sum + i*p10[pos]%MOD*tmp.cnt%MOD)%MOD;

        ans.sqsum=(ans.sqsum + i*i*p10[pos]%MOD*p10[pos]%MOD*tmp.cnt%MOD + 2*i*p10[pos]%MOD*tmp.sum%MOD + tmp.sqsum)%MOD;

    }

    if(!limit) dp[pos][mod]=ans;

    return ans;

}

ll solve(int n)

{

    int len=0;

    while(n){

        bit[len++]=n%10;

        n/=10;

    }

    return dfs(len-1,0,1).sqsum;

}

int main()

{

    init();

    int T;

    scanf("%d",&T);

    while(T--)

    {

        memset(dp,-1,sizeof(dp));

        scanf("%d%d%d",&l,&r,&k);

        printf("%lld\n",(solve(r)-solve(l-1)+MOD)%MOD); 

    }

    return 0;

}