HDU4722:Good Numbers(数位DP)

Problem Description

If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number.
You are required to count the number of good numbers in the range from A to B, inclusive.

 

Input

The first line has a number T (T <= 10000) , indicating the number of test cases.
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 10 ¹⁸).

 

Output

For test case X, output "Case #X: " first, then output the number of good numbers in a single line.

 

Sample Input

2 1 10 1 20

 

Sample Output

Case #1: 0 Case #2: 1

Hint

The answer maybe very large, we recommend you to use long long instead of int.

 

题意：求出区间内每位之和能被10整除的数的个数

思路：这题是简单的数位DP，当然用规律也能破

 

#include 
#include 
#include 
using namespace std;

__int64 bit[20];
__int64 dp[20][10];
//dp[i][j]，长度为i的数与10模为j的个数
__int64 solve(__int64 n)
{
    __int64 ans;
    __int64 tem1 = n,len,sum,i,x,j,k;
    len = sum = ans = 0;
    memset(dp,0,sizeof(dp));
    while(tem1)
    {
        bit[++len] = tem1%10;
        tem1/=10;
    }
    for(i = 1; i<=len/2; i++)//按高位在头排列
    {
        int t;
        t = bit[i];
        bit[i] = bit[len-i+1];
        bit[len-i+1] = t;
    }
    x = 0;
    for(i = 1; i<=len; i++)//从最高位开始
    {
        for(j = 0; j<10; j++)//将高位的全部枚举一次
            for(k = 0; k<10; k++)
                dp[i][(j+k)%10]+=dp[i-1][j];
        for(j = 0; j

 

而且通过观察我们可以发现，每个数从0到该位数字中，每位和能被10整除的个数是该数字除以10，然后再枚举各位到界限的个数之和

于是可得以下代码

 

#include 
#include 
#include 
using namespace std;

__int64 solve(__int64 n)
{
    if(n<0)
        return 0;
    __int64 t = n/100,i,ans = 0,sum;
    ans = t*10;
    for(i = t*100;i<=n;i++)
    {
        sum = 0;
        __int64 tem = i;
        while(tem)
        {
            sum+=tem%10;
            tem/=10;
        }
        if(sum%10 == 0)
        ans++;
    }

    return ans;
}

int main()
{
    __int64 T,l,r,cas = 1,ans;
    scanf("%I64d",&T);
    while(T--)
    {
        scanf("%I64d%I64d",&l,&r);
        ans = solve(r)-solve(l-1);
        printf("Case #%I64d: %I64d\n",cas++,ans);
    }

    return 0;
}