1283. Dwarf

Time limit: 1.0 second
Memory limit: 64 MB
Venus dwarfs are rather unpleasant creatures: ugly, malicious, and mean-spirited. Maybe it’s because of hard living conditions on their planet… but the fact remains: each of them is ready to sell his own mother in order to save up his pot of gold and to preserve it to the end of his days.
The dwarfs are especially nervous about the Mercury leprechauns who are always glad to empty a dwarf’s pot and to fill it with solar dust instead of the gold. The dwarfs are weak-sighted and can’t distinguish dust from gold. That is why each dwarf once a year visits the Central Galaxy Bank (CGB), where experienced specialists authenticate the content of the pot taking a small commission for the job.
When the amount of gold in a pot becomes less than or equal to a certain level, the life of a dwarf has no sense anymore, so he clears the world of his wretched soul: with the remaining gold he buys in a zoo the largest Jupiter toad and creeps under it which results in crushing his chest.

Input

The input contains three integers separated with spaces. The first number is the amount of gold in a dwarf’s pot at the initial moment. The second number is the amount of gold at which the dwarf’s life becomes senseless. Both values are measured in grams and don't exceed 231 − 1. The third number is the CGB commission (from 1 to 100); this is the percentage of gold that is taken from the pot as a way of payment for the verification.

Output

The output should contain the number of years that is left to the dwarf.

Samples

input output
19 10 50
1
1000 1 1
688
Problem Author: Leonid Volkov (prepared by Ivan Dashkevich)
Problem Source: USU Personal Contest 2004
Difficulty: 129
 
题意:输入一个数n,m表示钱的数量和下限(达到下限也会死),一个数k从1到100,表示每年用掉的钱的百分比,然后问多少年死?
分析:暴力
 
 #include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
using namespace std;
typedef long long LL;
typedef double DB;
#define For(i, s, t) for(int i = (s); i <= (t); i++)
#define Ford(i, s, t) for(int i = (s); i >= (t); i--)
#define Rep(i, t) for(int i = (0); i < (t); i++)
#define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
#define rep(i, x, t) for(int i = (x); i < (t); i++)
#define MIT (2147483647)
#define INF (1000000001)
#define MLL (1000000000000000001LL)
#define sz(x) ((int) (x).size())
#define clr(x, y) memset(x, y, sizeof(x))
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define ft first
#define sd second
#define mk make_pair
inline void SetIO(string Name)
{
string Input = Name + ".in",
Output = Name + ".out";
freopen(Input.c_str(), "r", stdin),
freopen(Output.c_str(), "w", stdout);
} inline int Getint() {
int Ret = ;
char Ch = ' ';
bool Flag = ;
while(!(Ch >= '' && Ch <= ''))
{
if(Ch == '-') Flag ^= ;
Ch = getchar();
}
while(Ch >= '' && Ch <= '') {
Ret = Ret * + Ch - '';
Ch = getchar();
}
return Flag ? -Ret : Ret;
} const DB Eps = 1e-;
DB a, b, c;
LL Ans; inline void Input() { } inline void Solve() {
for(cin >> a >> b >> c; a > b + Eps; a -= a * c / 100.0, Ans++) ;
cout << Ans << endl;
} int main() {
#ifndef ONLINE_JUDGE
SetIO("A");
#endif
Input();
Solve();
return ;
}

ural 1283. Dwarf的更多相关文章

  1. ural 1243. Divorce of the Seven Dwarfs

    1243. Divorce of the Seven Dwarfs Time limit: 1.0 secondMemory limit: 64 MB After the Snow White wit ...

  2. URAL - 1243 - Divorce of the Seven Dwarfs (大数取模)

    1243. Divorce of the Seven Dwarfs Time limit: 1.0 second Memory limit: 64 MB After the Snow White wi ...

  3. dwarf tower

    dwarf tower(dwarf.cpp/c/pas)[问题描述]Vasya在玩一个叫做"Dwarf Tower"的游戏,这个游戏中有n个不同的物品,它们的编号为1到n.现在Va ...

  4. dwarf格式解析

    debug_line中包含的是地址和源文件行之间的关系 我今天想搞清楚的是文件的C代码和汇编代码之间的关系: 对这块之前一直是迷迷糊糊的,发现这个问题已经严重影响到bug的定位了. 之前感觉C和汇编不 ...

  5. 后缀数组 POJ 3974 Palindrome && URAL 1297 Palindrome

    题目链接 题意:求给定的字符串的最长回文子串 分析:做法是构造一个新的字符串是原字符串+反转后的原字符串(这样方便求两边回文的后缀的最长前缀),即newS = S + '$' + revS,枚举回文串 ...

  6. ural 2071. Juice Cocktails

    2071. Juice Cocktails Time limit: 1.0 secondMemory limit: 64 MB Once n Denchiks come to the bar and ...

  7. ural 2073. Log Files

    2073. Log Files Time limit: 1.0 secondMemory limit: 64 MB Nikolay has decided to become the best pro ...

  8. ural 2070. Interesting Numbers

    2070. Interesting Numbers Time limit: 2.0 secondMemory limit: 64 MB Nikolay and Asya investigate int ...

  9. ural 2069. Hard Rock

    2069. Hard Rock Time limit: 1.0 secondMemory limit: 64 MB Ilya is a frontman of the most famous rock ...

随机推荐

  1. Class Methods & Variables

    When calling an instance method like withdraw_securely, the syntax generally looks something like th ...

  2. ubuntu下sh文件使用

    可把shell命令批处理写进filename.sh文件 然后执行 chmod +x filename.sh 就可以执行./filename.sh了

  3. Subarray Sum & Maximum Size Subarray Sum Equals K

    Subarray Sum Given an integer array, find a subarray where the sum of numbers is zero. Your code sho ...

  4. 高效PHP开发注意事项

    2015年2月26日 17:23:26 http://www.open-open.com/lib/view/open1332904714233.html

  5. Java for LeetCode 076 Minimum Window Substring

    Given a string S and a string T, find the minimum window in S which will contain all the characters ...

  6. Java for LeetCode 031 Next Permutation

    Next Permutation Total Accepted: 33595 Total Submissions: 134095     Implement next permutation, whi ...

  7. 22.python笔记之web框架

    一.web框架本质 1.基于socket,自己处理请求 #!/usr/bin/env python3 #coding:utf8 import socket def handle_request(cli ...

  8. css3学习总结8--CSS3 3D转换

    3D 转换 1. rotateX() 2. rotateY() otateX() 方法 通过 rotateX() 方法,元素围绕其 X 轴以给定的度数进行旋转. 示例: div { transform ...

  9. 在竞赛ACM Java处理输入输出

    一.Java之ACM注意点 1. 类名称必须采用public class Main方式命名 2. 在有些OJ系统上,即便是输出的末尾多了一个“ ”,程序可能会输出错误,所以在我看来好多OJ系统做的是非 ...

  10. 学习配置vsftp 进行ftp文件的传输

    一. FTP 说明 linux 系统下常用的FTP 是vsftp, 即Very Security File Transfer Protocol. 还有一个是proftp(Profession ftp) ...