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：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #define LL _int64 6 #include<algorithm> 7 using namespace std; 8 LL cal(LL n) 9 { 10 LL num,i,j,ans,sum; 11 num=n/100; 12 ans=num*10; 13 for(i=num*100;i<=n;i++) 14 { 15 j=i; 16 sum=0; 17 while(j) 18 { 19 sum+=j%10; //将各个位相加 20 j/=10; 21 } 22 if(sum%10==0) 23 ans++; 24 } 25 return ans; 26 } 27 void Init() 28 { 29 LL l,r; 30 static int count=1; 31 scanf("%I64d%I64d",&l,&r); 32 LL temp=cal(r)-cal(l-1); 33 printf("Case #%d: %I64d ",count++,temp); 34 } 35 int main() 36 { 37 int t; 38 scanf("%d",&t); 39 while(t--) 40 { 41 Init(); 42 } 43 return 0; 44 }