NYOJ 698 A Coin Problem (斐波那契)

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.

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

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!