链接：click here
题意：
描述
One day,Jiameier is tidying up the room,and find some coins. Then she throws the coin to play.Suddenly,she thinks of a problem ,that if throw n times coin ,how many situations of no-continuous up of the coin. Hey,Let‘s solve the
problem.
-
输入
- The first of input is an integer T which stands for the number of test cases. For each test case, there is one integer n (1<=n<=1000000000) in a line, indicate the times of throwing coins.
-
输出
- The number of all situations of no-continuous up of the coin, moduled by 10000.
-
样例输入
-
3
1
2
3
-
样例输出
-
2
3
5
-
来源
- SCU Programming Contest 2011
Preliminary
思路：大数的斐波那契。
代码：
#include <stdio.h>
int main()
{
int i;
int a[15010];
a[0]=0,a[1]=2,a[2]=3;
for(i=3;i<15010;i++)
a[i]=(a[i-1]+a[i-2])%10000;
int n;
scanf("%d",&n);
while(n--)
{
int m;
scanf("%d",&m);
printf("%d\n",a[m%15000]%10000);
}
return 0;
}
When you want to give up, think of why you persist until now!
NYOJ 698 A Coin Problem (斐波那契)
原文：http://blog.csdn.net/u013050857/article/details/43919257
