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
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1 #include<iostream>
2 #include<cstring>
3 #include<cstdlib>
4 #include<cstdio>
5 #include<algorithm>
6 #include<cmath>
7 #include<queue>
8 #include<map>
9
10 #define N 1505
11 #define M 15
12 #define mod 1000000007
13 #define mod2 100000000
14 #define ll long long
15 #define maxi(a,b) (a)>(b)? (a) : (b)
16 #define mini(a,b) (a)<(b)? (a) : (b)
17
18 using namespace std;
19
20 ll x,k,i,tmp;
21 int cnt;
22 /*
23 void ini()
24 {
25 int i,k,j;
26 for(i=1;i<=15;i++){
27 for(k=1;k<=15;k++){
28 int x=i;
29 for(j=1;j<=k;j++){
30 while(x%j!=0){
31 x++;
32 }
33 }
34 printf(" i=%d k=%d x=%d\n",i,k,x);
35 }
36 }
37 }*/
38
39 int main()
40 {
41 //ini();
42 //freopen("data.in","r",stdin);
43 //scanf("%d",&T);
44 //for(int cnt=1;cnt<=T;cnt++)
45 //while(T--)
46 cnt=1;
47 while(scanf("%I64d%I64d",&x,&k)!=EOF)
48 {
49 if(x==0 && k==0) break;
50 printf("Case #%d: ",cnt);cnt++;
51 tmp=-1;
52 for(i=1;i<=k;i++){
53 if(x%i!=0){
54 x=(x/i+1)*i;
55
56 }
57 if(x%i==0 && (x/i)==i-1){
58 //printf(" i=%I64d i-1=%I64d x=%I64d\n",i,x/(i),x);
59 tmp=x/i;
60 break;
61 }
62 }
63 if(tmp==-1){
64 printf("%I64d\n",x);
65 }
66 else printf("%I64d\n",tmp*(k-i)+x);
67 }
68
69
70
71 return 0;
72 }