Uva 11361 Investigating Div-Sum Property 解题报告（递推）

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Next generation contest – 4

Time Limit – 4 secs

Investigating Div-Sum Property

 

 

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

Output for Sample Input

3

1 20 1

1 20 2

1 1000 4

20

5

64

 

    解题报告：刘汝佳老师书上的题目。显然，这种题目一般都很恶心。思路可能并不难，但是代码实现却是比较复杂，码力强才能敲得又快又好（本人需加强……）。

    代码如下：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int k;int tenPow[11];int top;int digital[11];int mod(int n){    return (n%k+k)%k;}void decomposition(int a){    top=0;    while(a)    {        digital[top++] = a%10;        a/=10;    }}int matrix[11][90][90];void calMatrix(int n){    memset(matrix, 0, sizeof(matrix));    for(int i=0;i<10;i++)        matrix[0][i%k][i%k]++;    for(int i=1;i<n;i++) for(int j=0;j<10;j++)    {        for(int a=0;a<k;a++) for(int b=0;b<k;b++)        {            matrix[i][mod(a+j)][mod(b+tenPow[i]%k*j)]+=matrix[i-1][a][b];        }    }}int calResult(){    int front=0;    int res=0;    int sum[12];    memset(sum,0,sizeof(sum));    for(int i=top-1;i>=0;i--)        sum[i] = sum[i+1]+digital[i];    for(int i=top-1;i>0;i--)    {        for(int j=0;j<digital[i];j++)        {            res+=matrix[i-1][mod(-sum[i+1]-j)][mod(-tenPow[i]%k*front)];            front++;        }        front = (front*10)%k;    }    for(int j=0;j<=digital[0];j++) if(mod(front+j)==0 && mod(sum[1]+j)==0)        res++;    return res;}void work(){    int a,b;    scanf("%d%d%d",&a,&b,&k);    if(k>=90)    {        puts("0");        return;    }    decomposition(b);    calMatrix(top-1);    int ans = calResult();    decomposition(a-1);    ans -= calResult();    printf("%d\n", ans);}void init(){    tenPow[0]=1;    for(int i=1;i<10;i++)        tenPow[i] = tenPow[i-1]*10;}int main(){    init();    int T;    scanf("%d",&T);    while(T--)        work();}