Investigating Div-Sum Property UVA - 11361

最新推荐文章于 2018-11-29 22:08:00 发布

转载最新推荐文章于 2018-11-29 22:08:00 发布 · 57 阅读

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原文链接：http://www.cnblogs.com/l609929321/p/6895193.html

本文介绍了一种使用数位动态规划(DP)的方法来解决一类特定的整除问题：找出两个整数区间内既能被某数K整除且其各位数字之和也能被K整除的所有数的数量。通过实现数位DP算法，文章给出了解决方案并提供了一个具体的代码实例。

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;
}

转载于:https://www.cnblogs.com/l609929321/p/6895193.html