HDU 4722-Good Numbers

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. 

打表找规律~

发现每100位便可以找到10各个位置加起来是10的倍数的数字...只要判断a 和b 是100的几倍即可~~

CODE:

#include<iostream>#include<stdio.h>#include<string.h>#include<string>using namespace std;int main(){    int T;    scanf("%d",&T);    for(int tt=1;tt<=T;tt++)    {        long long a,b;        scanf("%I64d%I64d",&a,&b);        long long ans=(b/100-a/100)*10;        long long aa=a/100*100;        while(aa<a)        {//printf("1\n");            long long ta=aa;            long long t=0;            while(ta!=0)            {                t+=ta%10;                ta=ta/10;            }            if(t%10==0)            {                ans--;            }            aa++;        }        long long bb=b/100*100;        while(bb<=b)        {//printf("1\n");            long long tb=bb;            long long t=0;            while(tb!=0)            {                t+=tb%10;                tb=tb/10;            }            if(t%10==0)            {                ans++;            }            bb++;        }        printf("Case #%d: %I64d\n",tt,ans);    }    return 0;}