HDU 5620 KK's Steel <贪心思维>

leibniz_zhang

于 2016-07-24 16:17:35 发布

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解决一个数学问题：如何将长度为N（1≤N≤10^18）的钢材切割成尽可能多的不同长度段，且任意三段不能构成三角形。通过递归寻找斐波那契数列来确定最大切割数量。

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KK's Steel

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

Our lovely KK has a difficult mathematical problem:he has a

meters steel,he will cut it into steels as many as possible,and he doesn't want any two of them be the same length or any three of them can form a triangle.

Input

The first line of the input file contains an integer

, which indicates the number of test cases.

Each test case contains one line including a integer

,indicating the length of the steel.

Output

For each test case, output one line, an integer represent the maxiumum number of steels he can cut it into.

Sample Input

1
6

Sample Output

Hint

1+2+3=6 but 1+2=3 They are all different and cannot make a triangle.

Source

BestCoder Round #71 (div.2)

每次减一个尽可能小的木棒--

代码：

#include<cstdio>
#define LL long long
void solve()
{
    LL n;scanf("%lld",&n);
    if (n<3)
    {
        printf("1\n");
        return ;
    }
    LL shu[150];n-=3;
    shu[0]=1;shu[1]=2;
    for (int i=2;i;i++)
    {
        shu[i]=shu[i-1]+shu[i-2];
        if (n<shu[i])
        {
            printf("%d\n",i);
            return ;
        }
        n-=shu[i];
    }
}
int main()
{
    int t;scanf("%d",&t);
    while (t--)
        solve();
    return 0;
}