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
2013 ACM/ICPC Asia Regional Online —— Warmup2
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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\n",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 }