hdoj 5620 KK's Steel 【数学】

最新推荐文章于 2016-08-22 14:28:59 发布

笑着走完自己的路

最新推荐文章于 2016-08-22 14:28:59 发布

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本文介绍了一个关于如何将一根钢条切割成尽可能多的不等长部分，且任意三部分不能构成三角形的问题。通过使用斐波那契数列的方法，提供了一种有效的解决方案，并附带了完整的AC代码。

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KK's Steel

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 253 Accepted Submission(s): 120

Problem Description

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

N(1≤N≤1018) 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

T(1≤T≤10), which indicates the number of test cases.

Each test case contains one line including a integer

N(1≤N≤1018),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

3

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

题意：给你一根长n的木条，要求你截取若干个不等长的木条且任意三个都不能组成三角形，问最多截多少条。

思路：就是一个斐波那契，以前做过类似的题目。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 1000000
#define eps 1e-8
#define MAXN (10000+10)
#define MAXM (10000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
LL f[1001];
void getf()
{
    f[1] = 1; f[2] = 2;
    for(int i = 3; i <= 1000; i++)
        f[i] = f[i-1] + f[i-2];
}
int main()
{
    getf();
    int t; Ri(t);
    W(t)
    {
        LL n; scanf("%I64d", &n);
        LL sum = 0; int i;
        for(i = 1; i <= 1000; i++)
        {
            sum += f[i];
            if(sum >= n) break;
        }
        if(sum != n) i--;
        Pi(i);
    }
    return 0;
}