
第一篇.基本能力的编程实验
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POJ 1004 Financial Management
#include using namespace std;int main(){ double a[15]; double sum=0; int i; for(i=1;i<=12;i++) { cin>>a[i]; sum+=a[i]; } cout<<"$"<<sum/12.0<<endl;}原创 2014-08-06 09:49:31 · 705 阅读 · 0 评论 -
POJ 2136 Vertical Histogram(垂直直方图)
垂直柱状图显示字母出现的频率。原创 2014-07-13 12:14:12 · 1205 阅读 · 1 评论 -
POJ 1005 I Think I Need a Houseboat
I Think I Need a HouseboatTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 86518 Accepted: 37514DescriptionFred Mapper is considering purchasing some l原创 2014-08-06 10:20:27 · 638 阅读 · 0 评论 -
POJ 1019 Number Sequence
DescriptionA single positive integer i is given. Write a program to find the digit located in the position i in the sequence of number groups S1S2...Sk. Each group Sk consists of a sequence of pos原创 2014-07-15 11:36:01 · 878 阅读 · 0 评论 -
POJ 1338 Ugly Numbers
Ugly NumbersTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 20294 Accepted: 9018DescriptionUgly numbers are numbers whose only prime factors are 2, 3 or原创 2014-07-14 09:44:12 · 1089 阅读 · 2 评论 -
POJ 3006 Dirichlet's Theorem on Arithmetic Progressions
关于素数。原创 2014-08-06 10:41:09 · 835 阅读 · 0 评论 -
POJ 1003 Hangover
HangoverTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 101103 Accepted: 49150DescriptionHow far can you make a stack of cards overhang a table? If yo原创 2014-08-06 09:44:25 · 708 阅读 · 0 评论 -
POJ 3094 Quicksum
QuicksumTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13823 Accepted: 9634DescriptionA checksum is an algorithm that scans a packet of data and return原创 2014-08-06 10:45:01 · 665 阅读 · 0 评论 -
POJ 1833 排列
Description题目描述:大家知道,给出正整数n,则1到n这n个数可以构成n!种排列,把这些排列按照从小到大的顺序(字典顺序)列出,如n=3时,列出1 2 3,1 3 2,2 1 3,2 3 1,3 1 2,3 2 1六个排列。任务描述:给出某个排列,求出这个排列的下k个排列,如果遇到最后一个排列,则下1排列为第1个排列,即排列1 2 3…n。比如:n = 3,k原创 2014-07-14 19:55:03 · 1049 阅读 · 0 评论 -
POJ 2739 Sum of Consecutive Prime Numbers
Sum of Consecutive Prime NumbersTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 19035 Accepted: 10435DescriptionSome positive integers can be represente原创 2014-08-06 14:39:27 · 902 阅读 · 0 评论 -
POJ 1581 A Contesting Decision
结构体数组= =简单题原创 2014-08-07 16:49:24 · 975 阅读 · 0 评论 -
Ural 1068 sum
A -Sum(1.5.1)Crawling in process...Crawling failedTime Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uDescriptionYour task is to find the sum of all in原创 2014-08-07 09:07:24 · 867 阅读 · 0 评论 -
POJ 3299 Humidex
HumidexTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 16610 Accepted: 6041DescriptionAdapted from Wikipedia, the free encyclopediaThe humidex is a me原创 2014-08-06 10:58:15 · 737 阅读 · 0 评论 -
POJ 2196 Specialized Four-Digit Numbers
Specialized Four-Digit NumbersTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7310 Accepted: 5343DescriptionFind and list all four-digit numbers in deci原创 2014-08-06 10:51:13 · 809 阅读 · 0 评论 -
POJ 1552 Doubles
DoublesTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 19093 Accepted: 11027DescriptionAs part of an arithmetic competency program, your students will b原创 2014-08-06 14:16:34 · 646 阅读 · 0 评论