hdu 4722 Good Numbers ( 找规律 )

Good Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 549 Accepted Submission(s): 204

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

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

思路：

多写几个这样的数之后，你会发现这样的数每10个会出现一次，这个数/10+1就是第几个这样的数。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 1000005
using namespace std;

typedef long long ll;
ll n,m,ans,le,ri;

bool isok(ll u)
{
    ll v,t=0;
    while(u)
    {
        v=u%10;
        t+=v;
        u/=10;
    }
    if(t%10==0) return true ;
    return false ;
}
ll solve(ll u)
{
    int i,j;
    if(u==-1) return -1;
    while(!isok(u)) u--;
    return u/10;
}
int main()
{
    int i,j,t,test=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d",&le,&ri);
        ans=solve(ri)-solve(le-1);
        printf("Case #%d: %I64d\n",++test,ans);
    }
    return 0;
}

补充：软件开发 , C++ ,