ZOJ3498-Javabeans

Javabeans

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Javabeans are delicious. Javaman likes to eat javabeans very much.

Javaman has n boxes of javabeans. There are exactly i javabeans in the i-th box (i = 1, 2, 3,...n). Everyday Javaman chooses an integer x. He also chooses several boxes where the numbers of javabeans are all at least x. Then he eats x javabeans in each box he has just chosen. Javaman wants to eat all the javabeans up as soon as possible. So how many days it costs for him to eat all the javabeans?

Input

There are multiple test cases. The first line of input is an integer T ≈ 100 indicating the number of test cases.

Each test case is a line of a positive integer 0 < n < 2³¹.

Output

For each test case output the result in a single line.

Sample Input

41234

Sample Output

1223

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Author: CAO, Peng
Contest: The 8th Zhejiang Provincial Collegiate Programming Contest

题意：有n个盒子。第i个盒子有i个javabean，每天可以选择一个x，然后把盒子里有大于等于x个javabeans里吃掉x个，问最少需要几天才能吃完。
解题思路：n个盒子，每天吃（n+1）/2个，最多的盒子剩下n/2个

#include <iostream>#include <stdio.h>using namespace std;int main(){    int t,n;    cin>>t;    while(t--)    {        int sum=0;        scanf("%d",&n);        while(n>0)        {            n/=2;            sum++;        }        printf("%d\n",sum);    }    return 0;}