Problem H: Eat Candy

Problem H: Eat Candy

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 0   Solved: 0

Description

   There is a box with infinite volume. At first there are n candies in the box. Then every second you will eat some candies, left half of candies (round down) in the box. Then add k candies into the box. How many candies there are in the box after 109+7 seconds?

Input

   There are multiple test cases. In each test case, there are only one line contains two integers n,k(1≤n,k≤109+7)

Output

    For each test case, output the answer in one line.

Sample Input

4 5
2 3

Sample Output

9
5

HINT

In the first test case:

1st second, 4->2, 2+5 = 7

2nd second, 7->3, 3+5 = 8

3rd second, 8->4, 4+5 = 9

4th second, 9->4, 4+5 = 9

…

1000000007th            9

So there are 9 candies in the box after 1000000007 seconds.

解题思路:给你一个n和一个k,每秒执行一次n=n/2+k(四舍五入),问1000000007秒之后n为多少

因为操作固定,所以多次操作之后,n肯定会朝某个数单方向趋于一个稳定值,那我们只需要判断什么时候达到这个稳定值就可以了

简单一点的做法就是用STL里的map来标记已经出现过的数,当第二次访问到某一个数的时候就是那个稳定值

/*Sherlock and Watson and Adler*/
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<complex>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 100005;
const int M = 40;
const int inf = 100000000;
const int mod = 2009;
map<int,bool> m;
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        m.clear();
        while(!m[n])
        {
            m[n]=true;
            n/=2;
            n+=k;
        }
        printf("%d\n",n);
    }
    return 0;
}
 
/**************************************************************
    Problem: 19
    Language: C++
    Result: Accepted
    Time:17 ms
    Memory:1512 kb
****************************************************************/

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