ZOJ 3498 Javabeans(找规律）

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

4
1
2
3
4

Sample Output

1
2
2
3

题解：找规律

盒子数目：

1  2  3  4  5  6  7  8  9  10 11 12 13 14 15 16 17……

吃完要求的天数：

1  2  2  3  3  3  3  4  4   4   4   4   4   4   4   5  5……

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

#include <iostream>
#include<cstdio>
using namespace std;
int t;
long long n;
long long ans[];
int main()
{
    ans[]=;
    for(int i=;i<=;i++)
        ans[i]=ans[i-]*;   //之前爆int了，要用long long
    /*for(int i=1;i<=31;i++)
        printf("%d:%lld\n",i,ans[i]);*/
    while(~scanf("%d",&t))
    {
        for(int i=;i<=t;i++)
        {
            scanf("%lld",&n);
            for(int j=;j<=;j++)
            {
                if (ans[j]<n) continue;
                if (ans[j]==n) printf("%d\n",j+);
                else printf("%d\n",j);
                break;
            }
        }
    }
    return ;
}