(code jam)Problem C. Numbers

Problem

In this problem, you have to find the last three digits before the decimal point for the number (3 + √5)ⁿ.

For example, when n = 5, (3 + √5)⁵ = 3935.73982... The answer is 935.

For n = 2, (3 + √5)² = 27.4164079... The answer is 027.

Input

The first line of input gives the number of cases, T. T test cases follow, each on a separate line. Each test case contains one positive integer n.

Output

For each input case, you should output:

Case #X: Y

where X is the number of the test case and Y is the last three integer digits of the number (3 + √5)ⁿ. In case that number has fewer than three integer digits, add leading zeros so that your output contains exactly three digits.

Limits

1 <= T <= 100

Small dataset

2 <= n <= 30

Large dataset

2 <= n <= 2000000000

Sample

Input 
 
Output 
 2
5
2
Case #1: 935
Case #2: 027



题解：留着，等会写。先贴代码。不知道为什么提交不上去。运用了二项式原理。

因为是求（3+根号5）的n次方。我们可以添加一个（3-根号5）的n次方。其值远小于1，然后两者相加得s=a+b根号5.太麻烦了，不想写了。https://code.google.com/codejam/contest/32016/dashboard#s=a&a=2

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>#include <queue>#include <map>#include <stack>#include <list>#include <vector>using namespace std;#define DeBugstruct Mx{int a[3][3];}f,k;void Init(){k.a[1][1]=1;k.a[2][1]=0;k.a[1][2]=0;k.a[2][2]=1;}Mx mul(Mx x,Mx y){Mx z;int i,j,l;for (i=1;i<=2;i++)for (j=1;j<=2;j++){z.a[i][j]=0;for (l=1;l<=2;l++)z.a[i][j]=(z.a[i][j]+x.a[i][l]*y.a[l][j]) % 1000;}return z;}Mx ksm(Mx f,int n){if (n==1) return f;if (n % 2==0){Mx k1=ksm(f,n/2);return mul(k1,k1);}else return mul(f,ksm(f,n-1));}int main(){#ifdef DeBugfreopen("C-large-practice.in","r",stdin);freopen("output.out","w",stdout);#endifint T,n;scanf("%d",&T);f.a[1][1]=3;f.a[1][2]=5;f.a[2][1]=1;f.a[2][2]=3;for (int cas=1;cas<=T;cas++){scanf("%d",&n);Init();Mx ans=ksm(f,n);printf("#Case %d: %03d\n",cas,(2*ans.a[1][1]+999) % 1000);}return 0;}