KK's Steel
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 350 Accepted Submission(s): 166
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.
要求：给一个长整形数n让你把n分成若干份，1、任意两份长度不能相等 2、任意三分不能组成三角形
分析：因为斐波那契数列中的数满足此要求，所以从这个方面入手，只要分成的数满足a+b<=c即可，所以我们在数列1 2 3 5 8 13 21 ......的数中选取，如果前i项的和等于n输出i如果前i项的和大于n输出i-1
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define MAX 100100
#define INF 0x3f3f3f
#define LL long long
using namespace std;
LL fb[10010];
LL f[1001];
void biao()
{
LL i,j;
fb[1]=1;
fb[2]=2;
for(i=3;i<120;i++)
fb[i]=fb[i-1]+fb[i-2];
f[1]=fb[1];
for(i=2;i<120;i++)
f[i]=f[i-1]+fb[i];
}
int main()
{
int t,i,j;
LL n;
biao();
scanf("%d",&t);
while(t--)
{
scanf("%lld",&n);
for(i=1;i<120;i++)
{
if(n==f[i])
{
printf("%d\n",i);
break;
}
if(n<f[i])
{
printf("%d\n",i-1);
break;
}
}
}
return 0;
}