Description A boolean matrix has the parity property when each row and each column has an even sum, i.e. contains an even number of bits which are set. Here's a 4 x 4 matrix which has the parity property: 1 0 1 0 0 0 0 0 1 1 1 1 0 1 0 1 The sums of t…
Error Correction Time Limit: 2 Seconds      Memory Limit: 65536 KB A boolean matrix has the parity property when each row and each column has an even sum, i.e. contains an even number of bits which are set. Here's a 4 x 4 matrix which has the parity…
LINK:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3778 题意:有n道菜,每道菜需要\(a_i\)道工序,有m个锅可以同时做不同菜,问最小时间. 思路:水题,但要注意最小需要的时间显然是最大的\(a_i\)值,剩下的就是求个均,求个余的事了. /** @Date : 2017-03-24-14.35 * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : http…
题目链接:http://poj.org/problem?id=3254 Time Limit: 2000MS Memory Limit: 65536K Description Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows…
Information Entropy Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge Information Theory is one of the most popular courses in Marjar University. In this course, there is an important chapter about information entropy. Entropy is t…
题意:给定 n 个炸弹的坐标和爆炸时间,问你能不能逃出去.如果能输出最短时间. 析:其实这个题并不难,只是当时没读懂,后来读懂后,很容易就AC了. 主要思路是这样的,先标记所有的炸弹的位置,和时间,在数组中标记就好,只要赋值给它的爆炸时间就好,注意如果有多个,要赋值最小的那个, 然后用BFS走就行了. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #…
Description There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only…
We define b is a Divisor of a number a if a is divisible by b. So, the divisors of 12 are 1, 2, 3, 4, 6, 12. So, 12 has 6 divisors. Now you have to order all the integers from 1 to 1000. x will come before y if 1)                  number of divisors…
一.Description A simple mathematical formula for e is e=Σ0<=i<=n1/i! where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n. Input No input Output Output the approximations o…
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1796 题意: 四个人玩游戏,已知三个人的输赢情况,求第四个人的输赢情况. 思路: 设第一个人赢的次数为x输的次数为y 则 x+a1+a2+a3=y+b1+b2+b3 x+y=a1+b1 所以 x=(2*b1+b2+b3-a2-a3)/2 y=a1+b1-x #include<iostream> #include<cstdio> using namespace…