标签:斐波那契
链接:click here
题意:
3 1 2 3
2 3 5
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)
标签:斐波那契
原文地址:http://blog.csdn.net/u013050857/article/details/43919257