题目链接:10339 - Watching Watches 题意:两个时钟,一个每天慢a秒,一个每天慢b秒,问两钟重新相遇的时刻 1圈有12 * 60 * 60秒,然后1圈 / abs(a - b),就可以求出多少天会相遇,然后就能求出A钟一共慢了多少秒,进而可以求出该时刻的时和分! 下面给出AC代码: #include <bits/stdc++.h> using namespace std; int k,m; int main() { while(~scanf("%d%d"…

It has been said that a watch that is stopped keeps better time than one that loses 1 second per day.The one that is stopped reads the correct time twice a day while the one that loses 1 second per dayis correct only once every 43,200 days. This maxi…

题目大意:有两个表,每天都会慢一点时间,给出每天慢得秒数,问下一次重合的时刻. 解题思路:时刻重合也就是说整整差了一周,一周是12小时,用12小时的秒数除以两个表的相差那就是需要多少天的时间后重合,知道了需要多少天*每一天某个表走的时间(注意要减去少走的时间)即使答案,化成时间的格式即可,注意0时显示成12时. #include<bits/stdc++.h> using namespace std; int main() { int m, c, h, t1, t2; double t3; lo…

题意:n条线段(n <= 100000) (L<=R <= 1e9) ,m组询问(m <= 100000) 每次询问一个点的覆盖范围的最大值.一个点x对于一条包括其的线段,覆盖范围为min(x-L,R-x) 思路:考虑将线段一份为二,对于左边的那部分,以右端点排序,然后 二分找到右端点恰好满足的那个点为id,那么接下来要做的就是就是在[id,n]这个范围内找到L最小的那个点,能够通过求前缀最大来得到.那么左边最大距离为  seg[id].L-preLMax[id],右边最大的求法也…

10339 - Watching Watches Time limit: 3.000 seconds It has been said that a watch that is stopped keeps better time than one that loses 1 second per day. The one that is stopped reads the correct time twice a day while the one that loses 1 second per…

UVa 1262  Password 题目: Password   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description  Shoulder-surfing is the behavior of intentionally and stealthily watching the screen of another person's electronic…

偶数时,中位数之间的数都是能够的(包含中位数) 奇数时,一定是中位数 推导请找初中老师 #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> using namespace std; int box[1000000]; int main() { //freopen("in","r"…

Password Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVA. Original ID: 1262 64-bit integer IO format: %lld      Java class name: Main pid=18400" class="btn" style="">Prev Submit showpid=18401"…

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UVA - 10564 Paths through the Hourglass 题意: 要求从第一层走到最下面一层,只能往左下或右下走 问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径. f[i][j][k]从下往上到第i层第j个和为k的方案数 上下转移不一样,分开处理 没必要判断走出沙漏 打印方案倒着找下去行了,尽量往左走   沙茶的忘注释掉文件WA好多次   #include <iostream> #include <cstdio> #include <a…