Problem H: Eat Candy

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本文探讨了一个有趣的算法问题EatCandy。该问题描述为在一无限体积盒子中，初始存放n颗糖果，每秒取出盒子中糖果的一半并向下取整，然后加入k颗新的糖果。文章提供了C++代码实现，通过迭代计算预测在10^9+7秒后盒子中剩余的糖果数量。

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Problem H: Eat Candy
Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 1336  Solved: 646
[Submit][Status][Web Board]
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 integersn,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.


来源： http://acm.hzau.edu.cn/problem.php?cid=1009&pid=7

///很容易发现200（可能更小）次后就固定了
#include <iostream>
#include <cstdio>
#include <map>
#include <string>
#include <algorithm>
#include <stdlib.h>
#include <cstring>
using namespace std;
int P[1000];
long long  Handle(long long n,long long k)
{
    long long C = n;
    for(int i=1;i<=200;i++) C = C/2 + k;
    return C;
}
int main()
{
    //freopen("D:\\test.txt","r",stdin);
    //freopen("D:\\tzzt.out","w",stdout);
    long long n,k;
    while(~scanf("%lld %lld",&n,&k)) printf("%lld\n",Handle(n,k));
    return 0;
}