第一道：

传送门：NYOJ 45 棋盘覆盖

2^k*2^k=4^k，其实就是大数计算，-1后取3的倍数。问题本源出自《计算机算法设计与分析》的棋盘覆盖问题，原始解法是分治法，递推出公式f(k) = f(k - 1) * 4 + 1，f(1)=1,再化解下就是4^0 + 4^1 + 4^2 + ... + 4^(n-1)。有兴趣可以google原问题，附上一个不错的问题链接：棋盘覆盖

思路用大数加模拟就很简单了

Java版：

import java.math.BigInteger;
import java.util.Scanner;class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);int T = in.nextInt();while(T-- >0) {              BigInteger a = new BigInteger("1");int n = in.nextInt();for(int i = 1; i<=n; i++) {a = a.multiply(BigInteger.valueOf(4));}BigInteger ans = a.subtract(BigInteger.valueOf(1)).divide(BigInteger.valueOf(3));System.out.println(ans);}}
}

C/C++:

#include <stdio.h>
#include <string.h>int a[50];
int b[100];void f()
{//大数*2 int i;for(i = 0; i < 50; i++){a[i] *= 2; }for(i = 0; i < 50 ; i++){a[i + 1] += a[i] / 10;a[i] %= 10;}
}void g()
{//大数减一，再除以三 、b[0] - 1; int i;int k = 0, f = 0;for(i = 99; i >= 0; i --){if(b[i] != 0)break;}/*for(int j = 0; j < 100; j++)printf("%d", b[j]);printf("\n");*/for(; i >= 0 ; i--){k = k * 10 + b[i];if(k / 3 || f){printf("%d", k / 3);k %= 3;f = 1;}}if(f == 0)printf("0");printf("\n");
}void f1()
{//大数的平方 int i, j;for(i = 0; i < 50; i++){for(j = 0; j < 50; j++){b[i + j] += a[i] * a[j];}}for(i = 0; i < 100; i++){b[i + 1] += b[i] / 10;b[i] %= 10;}
}int main (void)
{int m;scanf("%d", &m);while(m --){memset(a, 0, sizeof(a));memset(b, 0, sizeof(b));a[0] = 1;int n, i;scanf("%d", &n);for(i = 0; i < n; i++)f();/*for(i = 0; i < 50; i++)printf("%d", a[i]);printf("\n");*/f1();g();}return 0;
} 

网上其他人提出的万进制版本？

 /**   Author:        illuz <iilluzen[at]gmail.com>*   Blog:          http://blog.csdn.net/hcbbt*   File:          45.cpp*   Lauguage:      C/C++*   Create Date:   2013-08-31 19:39:38*   Descripton:    nyoj 45, cheesboard coverage, wan jin zhi */
#include <cstdio>const int MAXN = 100;
int a[MAXN];            // 万进制，每位存4进制int main() {int n, m, c;scanf("%d", &n);while (n--) {scanf("%d", &m);if (m == 1) {printf("1\n");continue;}a[0] = 1;int t = 0;         // 已经用到第几位for (int i = 1; i < m; i++) {c = 0;        // c为进位数for (int j = 0; j <= t; j++) {a[j] = a[j] * 4 + c;c = a[j] / 10000;a[j] %= 10000;}if (c != 0) {t++;a[t] = c;}a[0]++;}// 处理最后的进位，其实就是99999999这种情况，这里可以不用
//      for (int j = 0; j <= t; j++) {
//          c = a[j] / 10000;
//          a[j] %= 10000;
//      }
//      if (c != 0) {
//          t++;
//          a[t] = c;
//      }printf("%d", a[t]);for (int i = t - 1; i >= 0; i--)printf("%04d", a[i]);printf("\n");}return 0;
}