zoukankan      html  css  js  c++  java
  • Codeforces Round #275 (Div. 2) B. Friends and Presents 二分+数学

    8493833                 2014-10-31 08:41:26     njczy2010                     B - Friends and Presents                          GNU C++     Accepted 31 ms 4 KB
    B. Friends and Presents
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.

    In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.

    Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.

    A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.

    Input

    The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.

    Output

    Print a single integer — the answer to the problem.

    Sample test(s)
    Input
    3 1 2 3
    Output
    5
    Input
    1 3 2 3
    Output
    4
    Note

    In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.

    In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.

      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 3000
     14 #define M 1005
     15 #define mod 1000000007
     16 //#define p 10000007
     17 #define mod2 100000000
     18 #define ll long long
     19 #define LL long long
     20 #define maxi(a,b) (a)>(b)? (a) : (b)
     21 #define mini(a,b) (a)<(b)? (a) : (b)
     22 
     23 using namespace std;
     24 
     25 ll cnt1,cnt2,x,y;
     26 ll c;
     27 ll tot;
     28 
     29 void ini()
     30 {
     31     tot=cnt1+cnt2;
     32 }
     33 
     34 ll gcd(ll a,ll b)
     35 {
     36     if(b==0){
     37         return a;
     38     }
     39     else
     40         return gcd(b,a%b);
     41 }
     42 
     43 ll lcm(ll a,ll b)
     44 {
     45     return a/gcd(a,b)*b;
     46 }
     47 
     48 ll ok(ll m)
     49 {
     50     ll c1,c2,c3;
     51     c1=m-m/x;
     52     c2=m-m/y;
     53     c3=m-m/x-m/y+m/c;
     54     //printf("   c1=%I64d c2=%I64d c3=%I64d c=%I64d m=%I64d
    ",c1,c2,c3,c,m);
     55     if(c1>=cnt1 && c2>=cnt2 && c1+c2-c3>=tot){
     56         return 1;
     57     }
     58     else{
     59         return 0;
     60     }
     61 }
     62 
     63 void solve()
     64 {
     65     c=lcm(x,y);
     66    // printf("  c=%I64d
    ",c);
     67     ll l,r,m;
     68     l=1,r=20000000009;
     69     m=(l+r)/2;
     70     while(l<r)
     71     {
     72       //  printf(" l=%I64d r=%I64d m=%I64d
    ",l,r,m);
     73         if(ok(m)==1){
     74             r=m;
     75         }
     76         else{
     77             l=m+1;
     78         }
     79         m=(l+r)/2;
     80     }
     81    // printf(" l=%I64d r=%I64d m=%I64d
    ",l,r,m);
     82     printf("%I64d
    ",m);
     83 }
     84 
     85 void out()
     86 {
     87 
     88 }
     89 
     90 int main()
     91 {
     92     //freopen("data.in","r",stdin);
     93     //freopen("data.out","w",stdout);
     94    // scanf("%d",&T);
     95    // for(int ccnt=1;ccnt<=T;ccnt++)
     96    // while(T--)
     97     while(scanf("%I64d%I64d%I64d%I64d",&cnt1,&cnt2,&x,&y)!=EOF)
     98     {
     99        // if(n==0 && m==0 ) break;
    100         //printf("Case %d: ",ccnt);
    101         ini();
    102         solve();
    103         out();
    104     }
    105 
    106     return 0;
    107 }
  • 相关阅读:
    文件夹选项查看显示所有文件和文件夹..确定后隐藏文件依然不能显示,.
    加速电脑启动,给电脑瘦身
    进行性肌营养不良症的治疗
    可变量程的直流电压表
    数字图象处理课件 下载
    两顾高楼
    技巧心得:给拥有Google AdSense 帐户 朋友的一点忠告
    进行性肌营养不良研究又有新的发现
    电子通讯系统 >> BAS系统在地铁环境控制中的应用及实现
    J2ME游戏开发中时钟的简单实现
  • 原文地址:https://www.cnblogs.com/njczy2010/p/4064882.html
Copyright © 2011-2022 走看看