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.
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).
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
数论,找规律
Hint
The answer maybe very large, we recommend you to use long long instead of int.#include<cstdio> __int64 sum(__int64 b) { int flag=0; __int64 s=b,sb; __int64 sum=0; while(s!=0) { sum+=s%10; s/=10; } sum=sum-b%10; sum%=10; if(sum==0 || sum+b%10>=10) flag=0; else flag=1; sb=b/10-flag; return sb; } int main() { __int64 a,b; int t,step=1,k; scanf("%d",&t); while(t--) { k=0; scanf("%I64d%I64d",&a,&b); if(a==0) k=1; printf("Case #%d: %I64d\n",step++,sum(b)-sum(a-1)+k); } return 0; }