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

21 101 20

 

Sample Output

Case #1: 0Case #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 <stdio.h>#include <string.h>#include <algorithm>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<bit[i]; j++)//枚举该位的数字到界限            dp[i][(x+j)%10]++;        x = (x+bit[i])%10;    }    if(!x)        dp[len][0]++;    return dp[len][0];}int main(){    __int64 T,l,r,cas = 1;    __int64 ans;    scanf("%I64d",&T);    while(T--)    {        ans = 0;        scanf("%I64d%I64d",&l,&r);        ans = solve(r)-solve(l-1);        printf("Case #%I64d: %I64d\n",cas++,ans);    }    return 0;}

 

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 <stdio.h>#include <string.h>#include <algorithm>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;}

 

           

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