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
",i,k,x);
        }
    }
}*/

int main()
{
    //ini();
    //freopen("data.in","r",stdin);
    //scanf("%d",&T);
    //for(int cnt=1;cnt<=T;cnt++)
    //while(T--)
    cnt=1;
    while(scanf("%I64d%I64d",&x,&k)!=EOF)
    {
        if(x==0 && k==0) break;
        printf("Case #%d: ",cnt);cnt++;
        tmp=-1;
        for(i=1;i<=k;i++){
            if(x%i!=0){
                x=(x/i+1)*i;
                
            }
            if(x%i==0 && (x/i)==i-1){
                    //printf(" i=%I64d i-1=%I64d x=%I64d
",i,x/(i),x);
                    tmp=x/i;
                    break;
                }
        }
        if(tmp==-1){
            printf("%I64d
",x);
        }
        else printf("%I64d
",tmp*(k-i)+x);
    }



    return 0;
}