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Poor Warehouse Keeper

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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:
bubuko.com,布布扣
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:
bubuko.com,布布扣
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:
bubuko.com,布布扣
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:
bubuko.com,布布扣
At that moment, the total price is exact 1.0. Jenny expects a final screen in form of:
bubuko.com,布布扣
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 }

 

Poor Warehouse Keeper

标签:des   style   blog   http   color   io   os   ar   java   

原文地址:http://www.cnblogs.com/njczy2010/p/4028399.html

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