返回信息流Another kind of Fibonacci
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
As we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , S(N) = A(0)2 +A(1)2+……+A(n)2.
Input
There are several test cases.
Each test case will contain three integers , N, X , Y .
N : 2<= N <= 231 – 1
X : 2<= X <= 231– 1
Y : 2<= Y <= 231 – 1
Output
For each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.
Sample Input
2 1 1
3 2 3
Sample Output
6
196
代码见下~
这是一条镜像帖。来源：北邮人论坛 / cpp / #35711同步于 2010/2/6
该镜像源已超过 30 天没有更新，可能在源站已被删除。
CPP机器人发帖
[求助]Fibonacci Again and ...
Rvtea
2010/2/6镜像同步4 回复
订阅后，新回复会通过你的通知中心匿名送达。
4 条回复
#include <stdio.h>
void main()
{
long x,y,a0=1,a1=1,sum,n,p=2;
while(scanf("%ld%ld%ld",&n,&x,&y)==3)
{
if(n>=2&&x>=2&&y>=2)
{
n--;
while(n)
{
sum=x*a1+y*a0;
p+=sum*sum;
a0=a1;
a1=sum;
n--;
}
}
printf("%ld\n",p%10007);
a0=1;
a1=1;
sum=0;
p=2;
}
}
显示的是“time limited exceeded”~这是什么原因？