zoj 2679 Old Bill
Old Bill
Time Limit: 2 Seconds Memory Limit: 65536 KB
Among grandfather��s papers a bill was found:
72 turkeys $_679_
The first and the last digits of the number that obviously represented the total price of those turkeys are replaced here by blanks (denoted _), for they are faded and are now illegible. What are the two faded digits and what was the price of one turkey?
We want to write a program that solves a general version of the above problem:
N turkeys $_XYZ_
The total number of turkeys, N, is between 1 and 99, including both. The total price originally consisted of five digits, but we can see only the three digits in the middle. We assume that the first digit is nonzero, that the price of one turkey is an integer number of dollars, and that all the turkeys cost the same price.
Given N, X, Y , and Z, write a program that guesses the two faded digits and the original price. In case that there is more than one candidate for the original price, the output should be the most expensive one. That is, the program is to report the two faded digits and the maximum price per turkey for the turkeys.
Input
The input consists of T test cases. The number of test cases (T) is given on the first line of the input file. The first line of each test case contains an integer N (0 < N < 100), which represents the number of turkeys. In the following line, there are the three decimal digits X, Y , and Z, separated by a space, of the original price $_XYZ_.
Output
For each test case, your program has to do the following. For a test case, there may be more than one candidate for the original price or there is none. In the latter case your program is to report 0. Otherwise, if there is more than one candidate for the original price, the program is to report the two faded digits and the maximum price per turkey for the turkeys. The following shows sample input and output for three test cases.
Sample Input
3
72
6 7 9
5
2 3 7
78
0 0 5
Sample Output
3 2 511
9 5 18475
0
#include <iostream>
#include <cstdio>
using namespace std;
int main(){
int x, y, z;
int i, j;
int n, t;
scanf("%d", &t);
while(t--){
scanf("%d %d %d %d", &n, &x, &y, &z);
for(i = ; i > ; i--){
for(j = ; j >= ; j--){
int p = i * + x * + y * + z * + j;
if(p / n * n == p){
printf("%d %d %d\n", i, j, p / n);
goto RL;
}
}
}
printf("0\n");
RL: continue;
}
return ;
}
zoj 2679 Old Bill的更多相关文章
- ZOJ 2679 Old Bill ||ZOJ 2952 Find All M^N Please 两题水题
2679:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1679 2952:http://acm.zju.edu.cn/onli ...
- ZOJ 2679 Old Bill(数学)
主题链接:problemCode=2679" target="_blank">http://acm.zju.edu.cn/onlinejudge/showProbl ...
- ZOJ People Counting
第十三届浙江省大学生程序设计竞赛 I 题, 一道模拟题. ZOJ 3944http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=394 ...
- ZOJ 3686 A Simple Tree Problem
A Simple Tree Problem Time Limit: 3 Seconds Memory Limit: 65536 KB Given a rooted tree, each no ...
- ZOJ Problem Set - 1394 Polar Explorer
这道题目还是简单的,但是自己WA了好几次,总结下: 1.对输入的总结,加上上次ZOJ Problem Set - 1334 Basically Speaking ac代码及总结这道题目的总结 题目要求 ...
- ZOJ Problem Set - 1392 The Hardest Problem Ever
放了一个长长的暑假,可能是这辈子最后一个这么长的暑假了吧,呵呵...今天来实验室了,先找了zoj上面简单的题目练练手直接贴代码了,不解释,就是一道简单的密文转换问题: #include <std ...
- ZOJ Problem Set - 1049 I Think I Need a Houseboat
这道题目说白了是一道平面几何的数学问题,重在理解题目的意思: 题目说,弗雷德想买地盖房养老,但是土地每年会被密西西比河淹掉一部分,而且经调查是以半圆形的方式淹没的,每年淹没50平方英里,以初始水岸线为 ...
- ZOJ Problem Set - 1006 Do the Untwist
今天在ZOJ上做了道很简单的题目是关于加密解密问题的,此题的关键点就在于求余的逆运算: 比如假设都是正整数 A=(B-C)%D 则 B - C = D*n + A 其中 A < D 移项 B = ...
- ZOJ Problem Set - 1001 A + B Problem
ZOJ ACM题集,编译环境VC6.0 #include <stdio.h> int main() { int a,b; while(scanf("%d%d",& ...
随机推荐
- Spring AOP初步总结(一)
学习AOP有段时间了,一直没空总结一下,导致有些知识点都遗忘了,之后会把以前学过的Spring核心相关的知识点总结一轮... 先大体介绍下Spring AOP的特点(均摘自"Spring i ...
- eclipse版本要求修改
eclipse要求打开的是java1.6,而安装的是java1.7,这个时候需要修改配置 找到JAVA的安装路径, 点击前往-电脑-资源库-Java-javaVCirtualMachines-...- ...
- shell中的-z
-z 字符串为"null",即是指字符串长度为零.
- UVA - 1252 Twenty Questions (状压dp)
状压dp,用s表示已经询问过的特征,a表示W具有的特征. 当满足条件的物体只有一个的时候就不用再猜测了.对于满足条件的物体个数可以预处理出来 转移的时候应该枚举询问的k,因为实际上要猜的物品是不确定的 ...
- chrom浏览器-F2使用方法一
由于F12是前端开发人员的利器,所以我自己也在不断摸索中,查看一些博客和资料后,自己总结了一下来帮助自己理解和记忆,也希望能帮到有需要的小伙伴,嘿嘿! 首先介绍Chrome开发者工具中,调试时使用最多 ...
- C-基础:atoi
C语言库函数名: atoi 功 能: 把字符串转换成整型数. 名字来源:ASCII to integer 的缩写. 原型: int atoi(const char *nptr); 函数说明: 参数np ...
- 几句话总结一个算法之RNN、LSTM和GRU
RNN 一般神经网络隐层的计算是h=g(w * x),其中g是激活函数,相比于一般神经网络,RNN需要考虑之前序列的信息,因此它的隐藏h的计算除了当前输入还要考虑上一个状态的隐藏,h=g(w*x+w' ...
- 第1节 flume:15、flume案例二,通过自定义拦截器实现数据的脱敏
1.7.flume案例二 案例需求: 在数据采集之后,通过flume的拦截器,实现不需要的数据过滤掉,并将指定的第一个字段进行加密,加密之后再往hdfs上面保存 原始数据与处理之后的数据对比 图一 ...
- Mac 下 Android Studio 安装
给大家介绍下 Mac Os 系统下的 Android Studio 的安装吧,二者步骤类似. 方法/步骤 1 首先下载 Mac 环境下的 Android Studio 的安装包,为 dmg 格式的 ...
- Mac屏幕亮度保存
关于保存屏幕亮度的方法,论坛上已有几种,搜索 NVRAM 会出来很多教程,在此不再详述,可以参考帖子http://www.idelta.info/archives/nvram_on_hackintos ...