hdu 4952

Number Transformation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 418    Accepted Submission(s): 201

Problem Description

   Teacher Mai has an integer x.
   He does the following operations k times. In the i-th operation, x becomes the least integer no less than x, which is the multiple of i.
   He wants to know what is the number x now.

 

Input

   There are multiple test cases, terminated by a line "0 0".
   For each test case, the only one line contains two integers x,k(1<=x<=10^10, 1<=k<=10^10).

 

Output

   For each test case, output one line "Case #k: x", where k is the case number counting from 1.

 

Sample Input

2520 10
2520 20
0 0

 

Sample Output

Case #1: 2520
Case #2: 2600

 

Source

2014 Multi-University Training Contest 8

 

Recommend

hujie

 

 

必须强调，不会的题一定不要偷懒，打表！！！！！！！！！！！！！！

 

 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<map>

 #define N 1505
 #define M 15
 #define mod 1000000007
 #define mod2 100000000
 #define ll long long
 #define maxi(a,b) (a)>(b)? (a) : (b)
 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 ll x,k,i,tmp;
 int cnt;
 /*
 void ini()
 {
     int i,k,j;
     for(i=1;i<=15;i++){
         for(k=1;k<=15;k++){
             int x=i;
             for(j=1;j<=k;j++){
                 while(x%j!=0){
                     x++;
                 }
             }
             printf(" i=%d k=%d x=%d\n",i,k,x);
         }
     }
 }*/

 int main()
 {
     //ini();
     //freopen("data.in","r",stdin);
     //scanf("%d",&T);
     //for(int cnt=1;cnt<=T;cnt++)
     //while(T--)
     cnt=;
     while(scanf("%I64d%I64d",&x,&k)!=EOF)
     {
         if(x== && k==) break;
         printf("Case #%d: ",cnt);cnt++;
         tmp=-;
         for(i=;i<=k;i++){
             if(x%i!=){
                 x=(x/i+)*i;

             }
             if(x%i== && (x/i)==i-){
                     //printf(" i=%I64d i-1=%I64d x=%I64d\n",i,x/(i),x);
                     tmp=x/i;
                     break;
                 }
         }
         if(tmp==-){
             printf("%I64d\n",x);
         }
         else printf("%I64d\n",tmp*(k-i)+x);
     }

     return ;
 }