hdu 4722 Good Numbers(找规律)

Good Numbers

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 <= 10¹⁸).

 

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.
 

 

ps：找规律中的大水题，可是我忘记了0也能整除10，以至于三个小时找了一个错误的规律~Orz

数位dp也可以做，暂时还没学0.0

代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>#include<stdlib.h>using namespace std;#define LL long long intLL solve(LL x){    LL sum=x/10;    LL first=x/10*10;    for(LL i=first; i<=x; i++)    {        LL m=0,k=i;;        while(k)        {            m+=(k%10);            k/=10;        }        if((m%10)==0)            return sum+1;    }    return sum;}int main(){    LL t,k=0;    scanf("%lld",&t);    while(k<t)    {        LL le,ri;        scanf("%lld%lld",&le,&ri);        le--;        LL sum1=solve(le);        if(le<0)            sum1=0;        LL sum2=solve(ri);        LL ans=abs(sum2-sum1);        printf("Case #%lld: ",++k);        printf("%lld\n",ans);    }    return 0;}

总结：对于极限数据的把握还不够，比赛的时候心态也不好，竟然把0能整除10给忘了，下次如果一道题错5次以上并且能确定思路正确的话，一定要多出几组数据区间的下限和上限数据，不能再犯同样的错误了