标签:hunnu
题意:
对于我们已知的斐波那契数列,现在给出一个n,要我们求出一个新的斐波那契数列起始项,使得n能在新的斐波那契数列中,要求起始项y最小
思路:
我们知道
a3 = a1+a2
a4 = a1+2*a2
a5 = 2*a1+3*a2
a6 = 3*a1+5*a2
可以得到
an = fib[n-2]*a[1]+fib[n-1]*a[2]
然后我们只需要枚举就可以了,猜测a[2]不会超过10W,然后居然也过了
队友用数学方法解的,那个才是正解,暴力是邪门歪道,囧
http://blog.csdn.net/u010579068/article/details/47452185
#include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <queue> #include <map> #include <set> #include <vector> #include <math.h> #include <bitset> #include <algorithm> #include <climits> using namespace std; #define ls 2*i #define rs 2*i+1 #define UP(i,x,y) for(i=x;i<=y;i++) #define DOWN(i,x,y) for(i=x;i>=y;i--) #define MEM(a,x) memset(a,x,sizeof(a)) #define W(a) while(a) #define gcd(a,b) __gcd(a,b) #define LL long long #define ULL unsigned long long #define N 100005 #define INF 0x3f3f3f3f #define EXP 1e-8 #define rank rank1 const int mod = 1000000007; LL fib[50]= {0,0,1}; LL n ; int main() { LL i,j,k; for(i = 3; i<50; i++) { fib[i] = fib[i-1]+fib[i-2]; } int t; scanf("%d",&t); while(t--) { scanf("%I64d",&n); int flag = 0; LL x,y,tz,ty; for(i = 45; i>2; i--) { for(ty=1; ty<=100000; ty++) { int tem = n-ty*fib[i]; if(ty*fib[i]+fib[i-1]>n) break; else if(tem%fib[i-1]==0&&tem/fib[i-1]<=ty) { x = tem/fib[i-1]; y = ty; flag = 1; break; } } if(flag) break; } printf("%I64d %I64d\n",x,y); } return 0; }
版权声明:本文为博主原创文章,未经博主允许不得转载。
BAPC2014 I&&HUNNU11589:Excellent Engineers
标签:hunnu
原文地址:http://blog.csdn.net/libin56842/article/details/47457687