题目链接 下午比赛的时候没有想出来,其实就是int型的数分为30个位,然后按照位来排列枚举. 题意:求n个数里面,取i个数异或的所有组合的和,i取1~n 分析: 将n个数拆成30位2进制,由于每个二进制位异或后相加和原来的数异或相加是一样的,所以只需要对每一位累加计算,用组合数学取数就行了,奇数个异或得1,偶数个异或得0,再乘以自己的二进制位值,复杂度O(30*n*n) #include <iostream> #include <cstdio> #include <cmath…

Wall Painting Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1026    Accepted Submission(s): 280 Problem Description Ms.Fang loves painting very much. She paints GFW(Great Funny Wall) every da…

Wall Painting Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4339    Accepted Submission(s): 1460 Problem Description Ms.Fang loves painting very much. She paints GFW(Great Funny Wall) every…

题意: 从给出的颜料中选出天数个,第一天选一个,第二天选二个... 例如:第二天从4个中选出两个,把这两个进行异或运算(xor)计入结果 对于每一天输出所有异或的和 $\sum_{i=1}^nC_{n}^{i}$ 思路: 0⊕0=0,1⊕0=1,0⊕1=1,1⊕1=0(同为0,异为1) 例如样例 4 1 2 10 1 这4个数的二进制表示分别为: 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 第一天: 分别选出 1, 2, 10 ,1 = 14 第二天: 从4个中选出2个进行异…

题目链接: Wall Painting Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2681    Accepted Submission(s): 857 Problem Description Ms.Fang loves painting very much. She paints GFW(Great Funny Wall) ev…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1348 Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3386    Accepted Submission(s): 968 Problem Description Once upon a time there was a gre…

传送门: POJ:点击打开链接 HDU:点击打开链接 以下是POJ上的题: Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 29121   Accepted: 9746 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's cast…

Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3139    Accepted Submission(s): 888 Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a w…

Problem Description Josh Lyman is a gifted painter. One of his great works is a glass painting. He creates some well-designed lines on one side of a thick and polygonal glass, and renders it by some special dyes. The most fantastic thing is that it c…

hdu 3037 Saving Beans 题目大意:n个数,和不大于m的情况,结果模掉p,p保证为素数. 解题思路:隔板法,C(nn+m)多选的一块保证了n个数的和小于等于m.可是n,m非常大,所以用到Lucas定理. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; ll n, m, p; ll qPow (ll a…