Cube Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/#/problem/show/80 Description As a student of the applied mathematics school of UESTC, WCM likes mathematics. Some day he found an interesting theorem that every positive inte…

HDU5053the Sum of Cube(水题) 题目链接 题目大意:给你L到N的范围,要求你求这个范围内的全部整数的立方和. 解题思路:注意不要用int的数相乘赋值给longlong的数,会溢出. 代码: #include <cstdio> #include <cstring> const int N = 10005; typedef long long ll; ll t[N]; void init () { for (ll i = 1; i <= N - 5; i++…

Cake Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/#/problem/show/48 Description ZC's birthday is comming, lxh gives a cake of n×m as present for his girlfriend.Now lxh wants to know how many second do you need to divide it to…

遮挡判断(shadow) Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/#/problem/show/26 Description 在一个广场上有一排沿着东西方向排列的石柱子,阳光从东边以一定的倾角射来(平行光).有的柱子可能被在他东边的高大的柱子的影子给完全遮挡住了.现在你要解决的问题是求出有多少柱子是没有被完全遮挡住的. 假设每个石柱子是一根细棒,而且都垂直于地面摆放. Input 输入包含多组数据…

8球胜负(eight) Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/#/problem/show/24 Description 8球是一种台球竞赛的规则.台面上有7个红球.7个黄球以及一个黑球,当然还有一个白球.对于本题,我们使用如下的简化规则:红.黄两名选手轮流用白球击打各自颜色的球,如果将该颜色的7个球全部打进,则这名选手可以打黑球,如果打进则算他胜.如果在打进自己颜色的所有球之前就把黑球打进,则…

Friendship of Frog Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5578 Description N frogs from different countries are standing in a line. Each country is represented by a lowercase letter. The distance betwee…

A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/problem/A Description Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maxi…

B. Plane of Tanks: Pro Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/175/problem/B Description Vasya has been playing Plane of Tanks with his friends the whole year. Now it is time to divide the participants into several…

BZOJ初级水题列表——献给那些想要进军BZOJ的OIers 代码长度解释一切! 注:以下代码描述均为C++ RunID User Problem Result Memory Time Code_Length 695765 Eolv 1000 Accepted 804 kb 0 ms 118 B 739478 Eolv 2463 Accepted 804 kb 0 ms 134 B 696662 Eolv 1968 Accepted 1272 kb 48 ms 137 B 739546 Eolv…

题目链接: Water problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 23    Accepted Submission(s): 14 Problem Description If the numbers 1 to 5 are written out in words: one, two, three, four, fi…

题目背景 做正经题是不可能做正经题的,这辈子都不可能做正经题的,毒瘤题又不会做毒瘤题,就是水题这种东西,才维持了蒟蒻的信心: 题目描述 这里有N+1 道水题,编号分别为0 ~N+1 ,每道水题都有它自己水的程度,具体为对应编号的每位上的数字的乘积,现在为了能够更好的刷水题,我们需要统计一点小东西,统计给出L,R,A,B四个值,具体统计的内容为编号在LL 到RR 这段区间中水的程度在A 到B 之间的水题的编号和,现在请你编写一个程序来完成这个统计并A掉这道水题,因为结果可能很大,请输出在对P 取模…

题目描述 LYK出了道水题. 这个水题是这样的:有两副牌,每副牌都有n张. 对于第一副牌的每张牌长和宽分别是xi和yi.对于第二副牌的每张牌长和宽分别是aj和bj.第一副牌的第i张牌能覆盖第二副牌的第j张牌当且仅当xi>=aj并且yi>=bj.(注意牌不能翻转)当然一张牌只能去覆盖最多一张牌,而不能覆盖好多张. LYK想让两副牌的各n张一一对应叠起来.它想知道第二副牌最多有几张能被第一副牌所覆盖. 输入格式(water.in) 第一行一个数n. 接下来n行,每行两个数xi,yi. 接下来n行,…

Flow Layout Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 3091   Accepted: 2148 Description A flow layout manager takes rectangular objects and places them in a rectangular window from left to right. If there isn't enough room in one r…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

CSU 1772 漫漫上学路 Time Limit: 1000MS   Memory Limit: 131072KB   64bit IO Format: %lld & %llu Submit Status Description 对于csuxushu来说,能够在CSU(California State University)上学是他一生的荣幸.CSU校园内的道路设计的十分精巧,由n+1条水平道路和n+1条竖直道路等距交错而成,充分体现了校园深厚的文化底蕴.然而不幸的是CS市每到夏季,天降大雨,…

1050: 写一个函数,使给定的一个二维数组(3×3)转置,即行列互换 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 154  Solved: 112[Submit][Status][Web Board] Description 写一个函数,使给定的一个二维数组(3×3)转置,即行列互换. Input 一个3x3的矩阵 Output 转置后的矩阵 Sample Input 1 2 3 4 5 6 7 8 9 Sample Output 1 4 7…

DP 水题 Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers. Write a program to find and print…

1195: 相信我这是水题 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 821  Solved: 219 Description GDUT中有个风云人物pigofzhou,是冰点奇迹队的主代码手,萌萌的师弟师妹们们经常会让pigofzhou帮他们Debug,因为师弟师妹们打代码使用编程语言的种类千奇百怪,pigofzhou为此很头疼.现在假设师弟师妹们只喜欢Java或者C++或者C,因为他希望所有人都学相同的编程语言,只有这样pigofzhou教…

1303: [CQOI2009]中位数图 Time Limit: 1 Sec  Memory Limit: 162 MBSubmit: 2340  Solved: 1464[Submit][Status][Discuss] Description 给出1~n的一个排列,统计该排列有多少个长度为奇数的连续子序列的中位数是b.中位数是指把所有元素从小到大排列后,位于中间的数. Input 第一行为两个正整数n和b ,第二行为1~n 的排列. Output 输出一个整数,即中位数为b的连续子序列个数.…

B - 大还是小? Time Limit:5000MS     Memory Limit:65535KB     64bit IO Format: Description 输入两个实数,判断第一个数大,第二个数大还是一样大.每个数的格式为: [整数部分].[小数部分] 简单起见,整数部分和小数部分都保证非空,且整数部分不会有前导 0.不过,小数部分的最 后可以有 0,因此 0.0 和 0.000 是一样大的. Input 输入包含不超过 20 组数据.每组数据包含一行,有两个实数(格式如前所述)…

ACM小白...非常费劲儿的学习中,我觉得目前我能做出来的都可以划分在水题的范围中...不断做,不断总结,随时更新 POJ: 1004 Financial Management 求平均值 杭电OJ: 1001  Sum Problem 相加1002  A + B Problem II 模拟人工加法-精度计算 1008 Elevator 很水,但是情况要考虑全面…

Codeforces Round #258 (Div. 2) Predict Outcome of the Game C. Predict Outcome of the Game time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are n games in a football tournament. Three…

Codeforces Round #258 (Div. 2) Sort the Array B. Sort the Array time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Being a programmer, you like arrays a lot. For your birthday, your friends ha…

Codeforces Round #258 (Div. 2) Game With Sticks A. Game With Sticks time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output After winning gold and silver in IOI 2014, Akshat and Malvika want to hav…

题意就是讲给出的字符串元音字母去掉,在每个辅音字母前加点,且小写输出...注意y也要去掉(以我英语挂科的水平也知道y是辅音字母)... 水题.. 直接上代码好了... #include <iostream> #include <stdio.h> #include <string.h> #define MAXN 100+10 using namespace std; char s[]={"aeiouAEIOUyY"}; int main(void) {…

1061: 从三个数中找出最大的数 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 154  Solved: 124[Submit][Status][Web Board] Description 定义一个带参的宏(或者模板函数),从三个数中找出最大的数. Input 3个短整型数,空格隔开 3个实数,空格隔开 3个长整数,空格隔开 Output 最大的数,对于实数保留2位小数. Sample Input 1 2 3 1.5 4.7 3.2 12345…

1059: 判别该年份是否闰年 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 222  Solved: 139[Submit][Status][Web Board] Description 给年份year,定义一个宏,以判别该年份是否闰年.提示:宏名可以定义为LEAP_YEAR,形参为y,既定义宏的形式为 #define LEAP_YEAR(y) (读者设计的字符串) Input 一个年份 Output 根据是否闰年输出,是输出"L",否…

1064: 输入三个字符串,按由小到大的顺序输出 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 471  Solved: 188[Submit][Status][Web Board] Description 输入三个字符串,按由小到大的顺序输出.分别使用指针和引用方式实现两个排序函数.在主函数中输入和输出数据. Input 3行字符串 Output 按照从小到大输出成3行.由指针方式实现. 按照从小到大输出成3行.由引用方式实现.   Sample…

LSS Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 128000/64000 KB (Java/Others) SubmitStatistic Next Problem Problem Description Time flies, four years passed, colleage is over. When I am about to leave, a xuemei ask me an ACM  problem, but…

DNA Sorting Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 80832   Accepted: 32533 Description One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instanc…