Investigating Div-Sum Property UVA - 11361

 

An integer is divisible by 3 if the sum of its digits is also divisible by 3. For example, 3702 is divisible
by 3 and 12(3+7+0+2) is also divisible by 3. This property also holds for the integer 9.
In this problem, we will investigate this property for other integers.
Input
The first line of input is an integer T (T < 100) that indicates the number of test cases. Each case is
a line containing 3 positive integers A, B and K. 1 ≤ A ≤ B < 2
31 and 0 < K < 10000.
Output
For each case, output the number of integers in the range [A, B] which is divisible by K and the sum
of its digits is also divisible by K.
Sample Input
3
1 20 1
1 20 2
1 1000 4
Sample Output
20
5
64

求n到m间有多少个数本身能整除k，各位相加和也能整除k，注意直接for循环用普通逻辑做会时间超限

需用到数位dp

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int a,b,MOD,T;
int dp[15][90][90],pow_10[15];
int f(int d,int m1,int m2)//数位dp
{
    if(dp[d][m1][m2]!=-1)
    {
        return dp[d][m1][m2];
    }
    dp[d][m1][m2]=0;
    for(int i=0; i<10; i++)
    {
        dp[d][m1][m2]+=f(d-1,(m1+i)%MOD,(m2+i*pow_10[d-1])%MOD);
    }
    return dp[d][m1][m2];
}
int calc(int x)//处理下每个数
{
    int len=0;
    if(!x)
    {
        len=1;
    }
    int t=x;
    while(t)
    {
        ++len;
        t/=10;
    }
    int res=0,LeftSide=0,SumDigits=0;
    for(int i=1; i<=len; i++)
    {
        while((ll)LeftSide+(ll)pow_10[len-i]-1ll<=(ll)x)
        {
            res+=f(len-i,SumDigits%MOD,LeftSide%MOD);
            LeftSide+=pow_10[len-i];
            ++SumDigits;
        }
    }
    return res;
}
int main()
{
    scanf("%d",&T);
    pow_10[0]=1;
    for(int i=1; i<=9; ++i)
    {
        pow_10[i]=pow_10[i-1]*10;//首先存下每位的值，如0对应1，1对应10
    }
    for(; T; --T)
    {
        scanf("%d%d%d",&a,&b,&MOD);
        if(MOD>82)//2的31次方各位数之和最大为81
        {
            puts("0");
            continue;
        }
        memset(dp,-1,sizeof(dp));
        for(int i=0; i<MOD; ++i)
        {
            for(int j=0; j<MOD; ++j)
            {
                dp[0][i][j]=0;
            }
        }
        dp[0][0][0]=1;
        printf("%d\n",calc(b)-calc(a-1));
    }
    return 0;
}