题目:计算以-2为基数的数的表示. 分析:数论.写出不同位数能表示的数字区间就能够找到规律. 长度为1:[1,1]: 长度为2:[-2,-1]: 长度为3:[2,5]: 观察发现,区间长度增长为1,2,4,8,..,2^k,而且奇偶间隔开: 这样能够按顺序找到相应的1的位置,每次减去相应的基底(-2^k)寻找下一个1的位置就可以. 说明:又是数论. #include <iostream> #include <cstdlib> #include <cstring> #in…

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

题目大意:十进制与十六进制之间的相互转换. #include <cstdio> int main() { #ifdef LOCAL freopen("in", "r", stdin); #endif ]; while (gets(str)) { int n; ] == 'x') { sscanf(str, "%x", &n); printf("%d\n", n); } else { sscanf(str,…

1.LA 5694 Adding New Machine 关键词:数据结构,线段树,扫描线(FIFO) #include <algorithm> #include <cstdio> #include <cstring> #include <string> #include <queue> #include <map> #include <set> #include <ctime> #include <cm…

The Tower of Babylon Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 437 Appoint description:  System Crawler  (2015-08-29) Description   Perhaps you have heard of the legend of the Tower of Babylon.…

Cellular Network Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1456 A cellular network is a radio network made up of a number of cells each served by a base station located in the cell. The base sta…

 Count the Trees  Another common social inability is known as ACM (Abnormally Compulsive Meditation). This psychological disorder is somewhat common among programmers. It can be described as the temporary (although frequent) loss of the faculty of sp…

题意: 有n个按先后顺序放置的不同大小不同位置的圆,求所有可见圆弧的长度. 分析: 这道题应该是大白书上例题 LA 2572 (求可见圆盘的数量) Kanazawa 的加强版,整体框架都差不多. 对于每个圆求出与其他圆相交的交点所对应的幅角(转化到[0, 2π]中),排个序,然后枚举每段弧的终点,如果不被后面放置的圆所覆盖则可见. 注意: 原本以为WA是精度问题,后来调大调小都一直WA,这里精度eps从1e-11到1e-13都没问题. 但是在判断弧的终点是否被圆所覆盖的时候要加上等号.也就是第6…

典型的两道矩阵快速幂求斐波那契数列 POJ 那是 默认a=0,b=1 UVA 一般情况是 斐波那契f(n)=(n-1)次幂情况下的(ans.m[0][0] * b + ans.m[0][1] * a): //POJ #include <cstdio> #include <iostream> using namespace std; ; struct matrix { ][]; }ans, base; matrix multi(matrix a, matrix b) { matrix…

UVA 11774 - Doom's Day 题目链接 题意:给定一个3^n*3^m的矩阵,要求每次按行优先取出,按列优先放回,问几次能回复原状 思路:没想到怎么推理,找规律答案是(n + m) / gcd(n, m),在topcoder上看到一个证明,例如以下: We can associate at each cell a base 3-number, the log3(R) most significant digits is the index of the row of the cel…