1133. Fibonacci Sequence
Time limit: 1.0 second
Memory limit: 64 MB
Memory limit: 64 MB
is an infinite sequence of integers that satisfies to Fibonacci conditionFi + 2 = Fi + 1 + Fi for any integer i. Write a program, which calculates the value of Fn for the given values of Fi and Fj.
Input
The input contains five integers in the following order: i, Fi, j, Fj, n.
−1000 ≤ i, j, n ≤ 1000, i ≠ j,
−2·109 ≤ Fk ≤ 2·109 (k = min(i, j, n), …, max(i, j, n)).
−1000 ≤ i, j, n ≤ 1000, i ≠ j,
−2·109 ≤ Fk ≤ 2·109 (k = min(i, j, n), …, max(i, j, n)).
Output
The output consists of a single integer, which is the value of Fn.
Sample
| input | output |
|---|---|
3 5 -1 4 5 |
12 |
1 #include <iostream> 2 using namespace std; 3 const int oo=2000000001; 4 int now; 5 __int64 i,fi,j,fj,n; 6 __int64 l=-oo,r=oo,mid,sum[3]; 7 int main() 8 { 9 cin >> i >> fi >> j >> fj >> n; 10 if (i > j) 11 { 12 swap(i,j); 13 swap(fi,fj); 14 }//进行调整，使得i比j考前 15 while (l <= r) 16 {//进行二分 17 mid=(l+r)>>1;//先取中间值，mid表示第i+1个数的值 18 sum[0]=fi; 19 sum[now=1]=mid; 20 int k=i+2,flag=-2; 21 while (k <= j) 22 {//计算加法运算次数 23 now=(now+1)%3; 24 ++k; 25 sum[now]=sum[(now+2)%3]+sum[(now+1)%3]; 26 if (sum[now] > oo) flag=1;//向上超界了用1进行标记 27 else if (sum[now] < (-oo)) flag=-1;//超出了最低限用-1进行标记 28 if (flag != -2) break;//超界了就不需要在进行运算了 29 } 30 if (flag == -2) 31 { 32 if (sum[now] > fj) flag=1;//没有超界但是大于fj表示mid开大了 33 else if (sum[now] < fj) flag=-1;//没有超界但是运算到了j小于fj表示mid开小了 34 else flag=0;//mid开正确了 35 } 36 if (flag == 1) r=mid-1;//开大了重置r 37 else if (flag == -1) l=mid+1;//开小了总之L 38 else if (flag == 0) break;//正好则跳出 39 } 40 sum[0]=fi; 41 sum[now=1]=mid;//进行模拟找到第n个数 42 if (i < n) 43 { 44 int k=i+2; 45 while (k <= n) 46 { 47 now=(now+1)%3; 48 ++k; 49 sum[now]=sum[(now+2)%3]+sum[(now+1)%3]; 50 } 51 } 52 else if (i > n) 53 { 54 int k=i-1; 55 while (k >= n) 56 { 57 now=(now+1)%3; 58 --k; 59 sum[now]=sum[(now+2)%3]-sum[(now+1)%3]; 60 } 61 } 62 else --now; 63 cout << sum[now] << endl; 64 return 0; 65 }