POJ-1004-Finanical Management】的更多相关文章

1.链接地址: http://poj.org/problem?id=1004 http://bailian.openjudge.cn/practice/1004 2.题目: 总时间限制: 1000ms 内存限制: 65536kB 描述 Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry has decid…
Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 126087   Accepted: 55836 Description Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry has deci…
一. 题目 Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 173910   Accepted: 65186 Description Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry ha…
求平均数,记得之前在杭电oj上做过一个求平均数的题目,结果因为题目是英文的,我就懵逼了 #include <stdio.h> int main() { ; double num; int i; ; i < ; ++i) { scanf("%lf",&num); sum += num; } printf("$%.2f",sum/12.0); }…
参考:https://www.cnblogs.com/BTMaster/p/3525008.html #include <iostream> #include <cstdio> #include <cstring> using namespace std; int main() { float a,sum; while (cin>>a)//要注意格式,能从文件读入多组数据! { sum=a; ;i<;i++) { cin>>a; sum+=…
Financial Management POJ - 1004 解题思路:水题. #include <iostream> #include <cstdio> #include <iomanip> #include <cmath> using namespace std; int main() { double sum=0.0; double now=0.0; while(cin>>now){ sum+=now; ;i<;i++){ cin&…
水两发去建模,晚饭吃跟没吃似的,吃完没感觉啊. ---------------------------分割线"水过....."------------------------------- POJ 3100 Root of the Problem http://poj.org/problem?id=3100 大意: 给定B和N,求一个数A使得A^N最接近B 太水了..... #include<cstdio> #include<cmath> int main()…
Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 165062   Accepted: 61316 Description Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry has deci…
1.题目描述: http://poj.org/problem?id=1004 2.解题过程 这个题目咋一看很简单,虽然最终要解出来的确也不难,但是还是稍微有些小把戏在里面,其中最大的把戏就是float 输入的精度问题,比如,一种最简单最直观的的解题方案如下: #include <stdlib.h> #include <stdio.h>   int main() { int sum = 0; float cb_this_month; for(int i=0;i<12;++i)…
简单题 #include <iostream> #include <stdio.h> #include <string.h> #include <stack> using namespace std; int main() { ]; ; ; i < ; ++i) { cin >> salary[i]; sum += salary[i]; } cout << << endl; ; }…
大意是,从输入六个数 .第i个数代表价值为i的有几个,平均分给两个人 ,明摆着的背包问题,本来以为把他转化为01背包.可是TLe,后来发现是12万的平方还多,所以妥妥的TLE,后来发现这是一个全然背包问题.然后即纠结了 ,没学过啊 .最后发现思想好i是蛮简单的,水水的A掉了.最后注意一下初始化问题和输入问题后就好了 #include <stdio.h> #include <string.h> int a[10]; int dp[120005]; int maxx(int a,int…
1000 A+B Problem 题目大意:输入两个数a和b,输出他们的和. 代码: #include <stdio.h> int main() { int a, b; while (scanf("%d%d" , &a, &b) != EOF) { printf("%d\n", a + b); } return 0; } 1004 Financial Management 题目大意:告诉你Larry的12个月的工资,求这12个月的工资的平…
POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value = acceptedNumber / avgAcceptRate + submittedNumber. 这里用到avgAcceptedRate的原因是考虑到通过的数量站的权重可能比提交的数量占更大的权重,所以给acceptedNumber乘上了一个因子. 当然计算value还有别的方法,比如POJ上…
以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight Moves1101 Gamblers1204 Additive equations 1221 Risk1230 Legendary Pokemon1249 Pushing Boxes 1364 Machine Schedule1368 BOAT1406 Jungle Roads1411 Annive…
感谢范意凯.陈申奥.庞可.杭业晟.王飞飏.周俊豪.沈逸轩等同学的收集整理.   题号:1003 Hangover求1/2+1/3+...1/n的和,问需多少项的和能超过给定的值 类似于Zerojudge a625 题号:1004 Financial Management 求12个实数的平均值   题号:1008 Maya Calendar 玛雅历 此题语言可选中文 PS:注意Haab历的日期是从0开始,而Tzolkin历的日期是从1开始   题号:1067 取石子游戏 此题语言为中文 PS:请百…
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…
题目:https://www.lydsy.com/JudgeOnline/problem.php?id=1004 http://poj.org/problem?id=2409 学习材料:https://www.cnblogs.com/nietzsche-oier/p/6883880.html https://files-cdn.cnblogs.com/files/HocRiser/Burnside.pdf bzoj 1004:这道题注意考虑单位元的那个置换. 然后用 polya 定理即可.不动点…
题目:https://www.lydsy.com/JudgeOnline/problem.php?id=1004 关于置换群:https://www.cnblogs.com/nietzsche-oier/p/6883880.html https://files-cdn.cnblogs.com/files/HocRiser/Burnside.pdf 原来 burnside 引理中的“不动点”是指一种不变化的方案啊: 这道题就用 burnside 引理,但给出的 m 个置换还不是置换群,需要再加一个…
Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 164431   Accepted: 61085 Description Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry has deci…
Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 182193   Accepted: 68783 Description Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry has deci…
此文来自北京邮电大学ACM-ICPC集训队 此50题在本博客均有代码,可以在左侧的搜索框中搜索题号查看代码. 以下是原文: POJ推荐50题1.标记“难”和“稍难”的题目可以看看,思考一下,不做要求,当然有能力的同学可以直接切掉.2.标记为A and B的题目是比较相似的题目,建议大家两个一起做,可以对比总结,且二者算作一个题目.3.列表中大约有70个题目.大家选做其中的50道,且每类题目有最低数量限制.4.这里不少题目在BUPT ACM FTP上面都有代码,请大家合理利用资源.5.50个题目要…
http://poj.org/problem?id=3041 #include <cstdio> #include <cstring> #include <vector> using namespace std; int e[1004][1004]; bool vis[1004]; vector <int > G[1004]; int n,m; int dfs(int s){ vis[s]=true; if(s==n*2+1){//printf("…
[POJ1004]Financial Management 试题描述 Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry has decided that he needs to grab hold of his financial portfolio and solve his financing pr…
poj 题目分类 按照ac的代码长度分类(主要参考最短代码和自己写的代码) 短代码:0.01K--0.50K:中短代码:0.51K--1.00K:中等代码量:1.01K--2.00K:长代码:2.01K以上. 短:1147.1163.1922.2211.2215.2229.2232.2234.2242.2245.2262.2301.2309.2313.2334.2346.2348.2350.2352.2381.2405.2406: 中短:1014.1281.1618.1928.1961.2054…
本文来自:http://www.cppblog.com/snowshine09/archive/2011/08/02/152272.spx 多版本的POJ分类 流传最广的一种分类: 初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. (5)构造法.(poj3295) (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法:…
这个题犯了一个小小的错误,double输出的时候用f才对,输入用lf即可.... http://poj.org/problem?id=1004 #include<stdio.h> int main() { ],sum,ave; int i; sum=; ; i<=; i++) { scanf("%lf",&a[i]); sum+=a[i]; } ave=sum/; printf("$%.2f",ave);//就是这里,所以这个代码用GCC和…
转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并查集======================================[HDU]1213   How Many Tables   基础并查集★1272   小希的迷宫   基础并查集★1325&&poj1308  Is It A Tree?   基础并查集★1856   More i…
Question:http://poj.org/problem?id=1004问题点:求平均值及格式化输出. Memory: 248K Time: 0MS Language: C++ Result: Accepted #include<iostream> #include<iomanip> using namespace std; int main(void) { double sum=0.0; ;i<=;i++) { double temp; cin>>temp…
1.链接地址: http://bailian.openjudge.cn/practice/2013 http://poj.org/problem?id=2013 2.题目: Symmetric Order Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 11793   Accepted: 7181 Description In your job at Albatross Circus Management (yes, it…
Potted Flower Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 3872   Accepted: 1446 Description The little cat takes over the management of a new park. There is a large circular statue in the center of the park, surrounded by N pots of f…