POJ.1552 Doubles(水) 题意分析 暴力 代码总览 #include <cstdio> #include <stdio.h> #define nmax 100 using namespace std; int a[nmax]; int n; int main() { //freopen("in.txt","r",stdin); while(true){ int t; scanf("%d",&t); i…

一.Description As part of an arithmetic competency program, your students will be given randomly generated lists of from 2 to 15 unique positive integers and asked to determine how many items in each list are twice some other item in the same list. Yo…

#include <stdio.h> #include <stdlib.h> ]; int cmp(const void *a, const void *b) { return *(int*)a - *(int*)b; } int main() { int n,i,j,result; ) { i = result = ; nums[i] = n; ++i; while(scanf("%d",&n) && n) { nums[i] = n;…

题目: 题意:题意:给出几个正数(2~15个),然后就是求有这些数字的2倍有没有和原先的正数相同的,求出有几个,没有就是0. 分析:水题.用数组解决,开一个数组存正数,另开一个数组用来存这些数的2倍,接着就搜索,然后注意一下结束的时候怎么处理就行. c普通方法: #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #inc…

#include<iostream> using namespace std; int main() { ]; int i,j; ; do{ sum=; ;num[i-]!=&&num[i-]!=-;i++) { cin>>num[i]; ;j<i;j++) ||num[i]/num[j]==0.5) sum++; } ]==-)break; cout<<sum<<endl; }); ; }…

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…

2992.357000 1000 A+B Problem1214.840000 1002 487-32791070.603000 1004 Financial Management880.192000 1003 Hangover792.762000 1001 Exponentiation752.486000 1006 Biorhythms705.902000 1005 I Think I Need a Houseboat686.540000 1011 Sticks647.566000 1007…

Doubles Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19954   Accepted: 11536 Description As part of an arithmetic competency program, your students will be given randomly generated lists of from 2 to 15 unique positive integers and as…

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…