G(x) 思路: 首先搞清楚每个位置上的值有什么意义, 如果第i位的值为1则 第i位与第i+1位不同,反之相同. 然后考虑s1和s2为什么会不一样, 这是由于x+1后比特位进位导致的,于是得出一个性质: 如果进位最高在第i位, 则  x的[0,i-1]位为1; x+1的[0,i-1]位为0; x/x+1 在[i+1,n]位置的值相同 同时可以确定G(x)大部分位置的情况. 最后大概是这样:   bn bn-1 ... bi+1 bi bi-1 bi-2 ... b1 x 1  0 0 ... 0…

题意:有一个字符串,若以"desu"结尾,则将末尾的"desu"替换为"nanodesu",否则在字符串末尾加上"nanodesu".分析:水题,直接模拟. AC代码: #include<stdio.h> #include<string.h> ]; int main() { int i,t; scanf("%d",&t); ;i<=t;i++) { scanf(&quo…

题意:定义f(i, j) = ai|ai+1|ai+2| ... | aj (| 指或运算),求有多少对f(i,j)<m.1 <= n <= 100000, 1 <= m <= 230 分析:直接暴力.先固定左端点i,枚举j,当f(i,j)>=m时剪枝就行了,理论上O(n^2)的复杂度是过不了的,但是数据水了,加了那个剪枝后速度奇快.... 网上有O(30*n)的算法,但是我看了好久都没能理解=_= AC代码: #include<stdio.h> ],b[]…

G(x) Time Limit: 2000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 184    Accepted Submission(s): 44 Problem Description For a binary number x with n digits (AnAn-1An-2 ... A2A1), we encode it as Where ""…

We Love MOE Girls Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 208    Accepted Submission(s): 157 Problem Description Chikami Nanako is a girl living in many different parallel worlds. In this…

A Bit Fun Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 423    Accepted Submission(s): 270 Problem Description There are n numbers in a array, as a0, a1 ... , an-1, and another number m. We de…

F(x) Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 382    Accepted Submission(s): 137 Problem Description For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x…

Minimum palindrome Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 260    Accepted Submission(s): 127 Problem Description Setting password is very important, especially when you have so many "in…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4738 题意:有n座岛和m条桥,每条桥上有w个兵守着,现在要派不少于守桥的士兵数的人去炸桥,只能炸一条桥,使得这n座岛不连通,求最少要派多少人去. 分析:只需要用Tarjan算法求出图中权值最小的那条桥就行了.但是这题有神坑. 第一坑:如果图不连通,不用派人去炸桥,直接输出0 第二坑:可能会有重边 第三坑:如果桥上没有士兵守着,那至少要派一个人去炸桥. 比赛的时候看完就想做了,但是图论太挫了,居然不会…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4737 题目大意:给定一系列数,F(i,j)表示对从ai到aj连续求或运算,(i<=j)求F(i,j)<=m的总数. 解题思路:或运算只会让值变大或保持不变.不断通过右移j来更新F(i,j),当aj>=m时所有的i<=j F(i,j)都大于等于m,因此从j后面继续扫数组:当aj<m而F(i,j)>=m时通过右移i来使F(i,j)<m.扫完整个数组即可. #include…