Good Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 673 Accepted Submission(s): 242
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
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.
Source
找规律......
代码：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#define LL _int64
#include<algorithm>
using namespace std;
LL cal(LL n)
{
LL num,i,j,ans,sum;
num=n/100;
ans=num*10;
for(i=num*100;i<=n;i++)
{
j=i;
sum=0;
while(j)
{
sum+=j%10; //将各个位相加
j/=10;
}
if(sum%10==0)
ans++;
}
return ans;
}
void Init()
{
LL l,r;
static int count=1;
scanf("%I64d%I64d",&l,&r);
LL temp=cal(r)-cal(l-1);
printf("Case #%d: %I64d\n",count++,temp);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
Init();
}
return 0;
}