KK's Steel

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1009    Accepted Submission(s): 487 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：给你一根长为n的木棒，怎你最多能把它分成多少段，并且要求任意两段长度不相等，任意三段不能组成三角形 思路：通过观察，只能分成 1 2 3 5 8 13 21 .......... 即斐波那契数列 #include<cstdio> #define LL long long int int main() { int t; LL n; LL f[100]; f[1] = 1; f[2] = 2; for(int i = 3 ; i <= 85 ; i++) { f[i] = f[i-1] + f[i-2]; } scanf("%d",&t); while(t--) { LL sum = 0; int falg = 0;- scanf("%I64d",&n); for(int i = 1 ; i <= 85 ; i++){ sum += f[i]; if(sum >= n) { falg = i; break; } } if(sum != n) falg--; printf("%d\n",falg); } return 0; }