[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
",(solve(r)-solve(l-1)+MOD)%MOD); 
    }
    return 0;
}