NYOJ 698 A Coin Problem (斐波那契)

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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

3123

样例输出

235

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!