Poor Warehouse Keeper
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1672 Accepted Submission(s): 463
Problem Description
Jenny is a warehouse keeper. He writes down the entry records everyday. The record is shown on a screen, as follow:
There are only two buttons on the screen. Pressing the button in the first line once increases the number on the first line by 1. The cost per unit remains untouched. For the screen above, after the button in the first line is pressed, the screen will be:
The exact total price is 7.5, but on the screen, only the integral part 7 is shown. Pressing the button in the second line once increases the number on the second line by 1. The number in the first line remains untouched. For the screen above, after the button in the second line is pressed, the screen will be:
Remember the exact total price is 8.5, but on the screen, only the integral part 8 is shown. A new record will be like the following:
At that moment, the total price is exact 1.0. Jenny expects a final screen in form of:
Where x and y are previously given. What’s the minimal number of pressing of buttons Jenny needs to achieve his goal?
Input
There are several (about 50, 000) test cases, please process till EOF. Each test case contains one line with two integers x(1 <= x <= 10) and y(1 <= y <= 109) separated by a single space - the expected number shown on the screen in the end.
Output
For each test case, print the minimal number of pressing of the buttons, or “-1”(without quotes) if there’s no way to achieve his goal.
Sample Input
1 1 3 8 9 31
Sample Output
0 5 11
Hint
For the second test case, one way to achieve is: (1, 1) -> (1, 2) -> (2, 4) -> (2, 5) -> (3, 7.5) -> (3, 8.5)
Source
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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 #include<set>
10 #include<string>
11 //#include<pair>
12
13 #define N 1005
14 #define M 1000005
15 #define mod 1000000007
16 #define inf 0x3f3f3f3f
17 //#define p 10000007
18 #define mod2 100000000
19 #define ll long long
20 #define LL long long
21 #define maxi(a,b) (a)>(b)? (a) : (b)
22 #define mini(a,b) (a)<(b)? (a) : (b)
23
24 using namespace std;
25
26 ll mul=3628800;
27 ll x,y;
28 ll ans;
29 ll tmi,tma;
30 ll dmi,dma;
31 ll dan;
32 ll num,tot;
33
34 void ini()
35 {
36 ans=0;
37 tmi=y*mul;
38 tma=(y+1)*mul;
39 dmi=tmi/x;
40 dma=tma/x;
41 num=1;
42 dan=mul;
43 tot=mul;
44 }
45
46
47 void solve()
48 {
49 ans=x-1;
50 ll k1,k2;
51 ll te;
52 if(dan>=dma){
53 ans=-1;return;
54 }
55 for(num=1;num<=x;num++){
56 te=mul/num;
57 if(dan>=dmi) break;
58 k1=(dmi-dan+te-1)/te;
59 k2=(dma-dan+te-1)/te;
60 // printf(" num=%I64d ans=%I64d dan=%I64d dmi=%I64d dma=%I64d te=%I64d k1=%I64d k2=%I64d\n",
61 // num,ans,dan,dmi,dma,te,k1,k2);
62 if(k1!=k2){
63 ans+=k1;
64 dan+=k1*mul/num;
65 return;
66 }
67 else{
68 dan+=(k1-1)*te;
69 ans+=(k1-1);
70 }
71 }
72 }
73
74 void out()
75 {
76 printf("%I64d\n",ans);
77 }
78
79 int main()
80 {
81 //freopen("data.in","r",stdin);
82 //freopen("data.out","w",stdout);
83 //scanf("%d",&T);
84 // for(int ccnt=1;ccnt<=T;ccnt++)
85 // while(T--)
86 while(scanf("%I64d%I64d",&x,&y)!=EOF)
87 {
88 //if(n==0 && k==0 ) break;
89 //printf("Case %d: ",ccnt);
90 ini();
91 solve();
92 out();
93 }
94
95 return 0;
96 }