Priest John's Busiest Day Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1420    Accepted Submission(s): 415 Problem Description John is the only priest in his town. October 26th is the John's…

Timer Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 445    Accepted Submission(s): 90 Problem Description Recently, some archaeologists discovered an ancient relic on a small island in the Pa…

A simple stone game                                                                                                       Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)                                             …

Destroying the bus stations                                                                                     Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)                                                       …

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5122 解题报告:定义一种排序算法,每一轮可以随机找一个数,把这个数与后面的比这个数小的交换,一直往后判断,直到后面没有比这个数小的,这样称为一轮,现在给定一个长度为n的序列,要你求,至少要经过多少轮,可以使这个序列成为有序的. 由于只能跟后面的数进行比较,所以我只要统计后面的数有比这个数小的数的个数就可以了.从后往前扫一遍,每次更新当前最小的. #include<cstdio> #include&…

虽然是一道还是算简单的DP,甚至不用滚动数组也能AC,数据量不算很大. 对于N个数,每个数只存在两个状态,取 和 不取. 容易得出状态转移方程: dp[i][j] = dp[i - 1][j ^ a[i]] + dp[i - 1][j]; dp[i][j] 的意思是,对于数列 中前 i 个数字,使得 XOR 和恰好为 j 的方案数 状态转移方程中的 dp[i - 1][j] 即表示当前这个数字不取, dp[i - 1][j ^ a[i]] 表示当前这个数字要取. 这道题还是要好好理解阿! sou…

题意:有n个数字,带入10000 - (100 - ai) ^ 2公式得到n个数,输出n个数中频率最大的数,如果有并列就按值从小到大都输出输出,如果频率相同的数字是全部的n个数,就输出Bad....题解:统计数字个数和频率,排序后输出. Sample Input36100 100 100 99 98 1016100 100 100 99 99 1016100 100 98 99 99 97 Sample OutputCase #1:10000Case #2:Bad MushroomCase #3…

两个圆环的内外径相同 给出内外径 和 两个圆心 求两个圆环相交的面积 画下图可以知道 就是两个大圆交-2*小圆与大圆交+2小圆交 Sample Input22 30 00 02 30 05 0 Sample OutputCase #1: 15.707963Case #2: 2.250778 # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # i…

把一个序列按从小到大排序 要执行多少次操作 只需要从右往左统计,并且不断更新最小值,若当前数为最小值,则将最小值更新为当前数,否则sum+1 Sample Input255 4 3 2 155 1 2 3 4 Sample OutputCase #1: 4Case #2: 1 # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include…

给出某个时刻对应的速度 求出相邻时刻的平均速度 输出最大值 Sample Input23 // n2 2 //t v1 13 430 31 52 0 Sample OutputCase #1: 2.00Case #2: 5.00 # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <string> # include…