ACdreamoj 1417 思维题

http://115.28.76.232/contest?cid=1128#problem-E

Problem Description

      Consider numbers from 1 to n.

      You have to find the smallest lexicographically number among them which is divisible by k.

Input

      Input file contains several test cases. Each test case consists of two integer numbers n and k on a line(1 ≤ n ≤ 1018,
1 ≤ k ≤ n).
      The last test case is followed by a line that contains two zeroes. This line must not be processed.

Output

      For each test case output one integer number — the smallest lexicographically number not exceeding n which is divisible by k.

Sample Input

2000 17
2000 20
2000 22
0 0

Sample Output

1003
100
1012

解题思路：

               找到n和m的位数x和y，找到位数分别为 y+1~x 的最小的一个满足%k==0的数，然后和k~10^y+1之间所有满足的数作比较，保留字典序最小的就好了。

/*
* this code is made by life4711
* Problem: 1417
* Verdict: Accepted
* Submission Date: 2014-10-04 15:32:42
* Time: 24MS
* Memory: 1676KB
*/
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;

typedef long long LL;

LL n,k,flag;
char c[25],s[25],st[25];

int main()
{
while(~scanf("%lld%lld",&n,&k))
{
if(n==0&&k==0)
break;
LL x=1;
LL y=1;
flag=0;
while(x<n)
{
if(x*10>k&&flag==0)
{
flag=1;
y=x*10;
}
x*=10;
}
//printf("%lld %lld\n",x,y);
st[0]='9'+1;
LL t,i;
flag=-1;
while(x>=k)
{
if(x%k!=0)
t=k-x%k;
else
t=0;
if(t+x>n)
{
x/=10;
continue;
}
//printf("x,t:[%d,%d]\n",x,t);
flag=t+x;
if(flag!=-1)
{
t=0;
while(flag>0)
{
s[t++]=flag%10+'0';
flag/=10;
}
s[t]=0;
for(int i=0; i<t; i++)
c[i]=s[t-1-i];
c[t]=0;
//printf("%s\n%s\n",s,c);
if(strcmp(st,c)>0)
strcpy(st,c);
}
x/=10;
}
for(LL i=k;i<y&&i<=n;i+=k)
{

// printf("草泥马\n");
if(i%k==0)
{
flag=i;
t=0;
while(flag>0)
{
s[t++]=flag%10+'0';
flag/=10;
}
s[t]=0;
for(int j=0; j<t; j++)
c[j]=s[t-1-j];
c[t]=0;
//printf("%s\n%s\n",s,c);
if(strcmp(st,c)>0)
strcpy(st,c);
}
}
printf("%s\n",st);
}
return 0;
}
/*
100002150155 1000000001

1000 100

1254 12

13535 2165

2163516 3216

32135 21321

125 120

*/