MBLHQ的博客

TilingTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8319 Accepted: 4012DescriptionIn how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles?

hdu--1047                                         Integer InquiryTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15521    Accepted Submi

HDU---1042                                           N!Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 65926    Accepted Submission(s):

/*大数加法的关键在于用字符型的数组解决问题；两个字符数组用于存储将要进行计算的数据，另一个整形数组或字符数组存储结果，例如定义：char a[],b[];int c[];则c[]=a[]-'0'+b[]-'0';接着就是处理进位问题*/ #include#include#define M 10002char a[M],b[M];int c[M];int main(){

/*求N的阶乘*/（杭电---------N！）代码如下：#include#include#define M 10000000//刚开始开了100000结果溢出了 int a[M];int main(){ int n,p,h,i,j; a[1]=1;p=1;h=0;//p表示每次运算时的位数，h表示进位 while(scanf("%d",&n)!=EOF) { me

/*大整数的乘法*/#include#include#define M 100000char a[M];char b[M];int c[M+10];int main(){ int i,j,k,n,l1,l2,m; scanf("%d",&n); getchar(); while(n--) { gets(a); gets(b); k=