SRM 398(1-250pt)
题意:有两个变量x和y,三种运算符+,*,-,组成等式"变量 运算符 变量 运算符 变量 运算符 变量",要求每个变量恰好出现两次,且等式的值为val的等式有多少个。注意不计算运算符的优先级,比如2+2*3先计算2+2,最终结果得12。
解法:暴力即可。O(2^4 * 3^3)。终于交到200pt了...
tag:brute-force
// BEGIN CUT HERE
/*
* Author: plum rain
* score :
*/
/* */
// END CUT HERE
#line 11 "CountExpressions.cpp"
#include <sstream>
#include <stdexcept>
#include <functional>
#include <iomanip>
#include <numeric>
#include <fstream>
#include <cctype>
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <set>
#include <queue>
#include <bitset>
#include <list>
#include <string>
#include <utility>
#include <map>
#include <ctime>
#include <stack> using namespace std; #define CLR(x) memset(x, 0, sizeof(x))
#define CLR1(x) memset(x, -1, sizeof(x))
#define PB push_back
#define SZ(v) ((int)(v).size())
#define ALL(t) t.begin(),t.end()
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false) typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef pair<int, int> pii;
typedef long long int64; const double eps = 1e-;
const double PI = atan(1.0)*;
const int maxint = ; int num(int x)
{
int ret = ;
while (x){
if (x & ) ++ ret;
x >>= ;
}
return ret;
} int gao(int s, int t, int x, int y)
{
int num = s & ? x : y, times = ;
s >>= ;
while (times < ){
int tmp = s & ? x : y;
int suan = t % ;
if (suan == ) num = tmp + num;
else if (suan == ) num = tmp * num;
else num = num - tmp; s >>= ; t /= ;
++ times;
}
return num;
} class CountExpressions
{
public:
int calcExpressions(int x, int y, int val){
int cnt = ;
int the = * * ;
for (int i = ; i < (<<); ++ i) if (num(i) == )
for (int j = ; j < the; ++ j)
if (gao(i, j, x, y) == val) ++ cnt;
return cnt;
} // BEGIN CUT HERE
public:
void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); if ((Case == -) || (Case == )) test_case_5(); }
private:
template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
void test_case_0() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, calcExpressions(Arg0, Arg1, Arg2)); }
void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, calcExpressions(Arg0, Arg1, Arg2)); }
void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, calcExpressions(Arg0, Arg1, Arg2)); }
void test_case_3() { int Arg0 = -; int Arg1 = ; int Arg2 = -; int Arg3 = ; verify_case(, Arg3, calcExpressions(Arg0, Arg1, Arg2)); }
void test_case_4() { int Arg0 = ; int Arg1 = -; int Arg2 = -; int Arg3 = ; verify_case(, Arg3, calcExpressions(Arg0, Arg1, Arg2)); }
void test_case_5() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, calcExpressions(Arg0, Arg1, Arg2)); } // END CUT HERE }; // BEGIN CUT HERE
int main()
{
// freopen( "a.out" , "w" , stdout );
CountExpressions ___test;
___test.run_test(-);
return ;
}
// END CUT HERE
SRM 398(1-250pt)的更多相关文章
- SRM593(1-250pt,500pt)
SRM 593 DIV1 250pt 题意:有如下图所示的平面,每个六边形有坐标.将其中一些六边形染色,要求有边相邻的两个六边形不能染同一种颜色.给定哪些六边形需要染色,问最少需要多少种颜色. 解法: ...
- SRM475 - SRM479(1-250pt,500pt)
SRM 475 DIV1 300pt 题意:玩游戏.给一个棋盘,它有1×n(1行n列,每列标号分别为0,1,2..n-1)的格子,每个格子里面可以放一个棋子,并且给定一个只含三个字母WBR,长度为n的 ...
- SRM468 - SRM469(1-250pt, 500pt)
SRM 468 DIV1 250pt 题意:给出字典,按照一定要求进行查找. 解法:模拟题,暴力即可. tag:water score: 0.... 这是第一次AC的代码: /* * Author: ...
- SRM470 - SRM474(1-250pt,500pt)(471-500pt为最短路,474-500pt未做)
SRM 470 DIV1 250pt 题意:有n个房间排成一排,相邻两个房间之间有一扇关闭着的门(共n-1扇),每个门上都标有‘A’-‘P’的大写字母.给定一个数n,表示第n个房间.有两个人John和 ...
- SRM 609(1-250pt, 1-500pt)
嗯....还是应该坚持写题解的好习惯啊... DIV1 250pt 这难度是回到srm 300+的250了嘛...略 // BEGIN CUT HERE /* * Author: plum rain ...
- SRM 442(1-250pt, 1-500pt)
DIV1 250pt 题意:将一个数表示成质因子相乘的形式,若乘式所含数字的个数为质数,则称A为underprime.比如12 = 2*2*3,则含3个数字,是underprime.求A, B之间un ...
- 【SRM】649 t2
题意 一个数列\(A\),数的范围均在\([0, 2^N-1]\)内,求一个\(B\),使得新生成的数列\(C\)中逆序对最多(\(C_i = A_i xor B\)),输出最多的逆序对.(\(|A| ...
- 记第一次TopCoder, 练习SRM 583 div2 250
今天第一次做topcoder,没有比赛,所以找的最新一期的SRM练习,做了第一道题. 题目大意是说 给一个数字字符串,任意交换两位,使数字变为最小,不能有前导0. 看到题目以后,先想到的找规律,发现要 ...
- SRM 513 2 1000CutTheNumbers(状态压缩)
SRM 513 2 1000CutTheNumbers Problem Statement Manao has a board filled with digits represented as St ...
随机推荐
- Hadoop的读写类调用关系_图示
- 线程同步(AutoResetEvent与ManualResetEvent)
前言 在我们编写多线程程序时,会遇到这样一个问题:在一个线程处理的过程中,需要等待另一个线程处理的结果才能继续往下执行.比如:有两个线程,一个用来接收Socket数据,另一个用来处理Socket数据, ...
- webViewDidFinishLoad 执行多次的问题
在做网页加载进度条的时候,发现UIWebViewDelegate中webViewDidFinishLoad方法会执行多次: - (void)webViewDidStartLoad:(UIWebView ...
- qrcode-php生成二维码
调用PHP QR Code非常简单,如下代码即可生成一张内容为"http://www.baidu.com"的二维码. include 'phpqrcode.php'; QRcode ...
- Python:运算符
#!/usr/bin/python3 #运算符 #算术运算符 print("算术运算符:","+ - * / % **(幂) //(取整)") #比较运算符 p ...
- java_设计模式_外观模式_Facade Pattern(2016-08-09)
外观模式/门面模式 1.概念 为子系统中的一组接口提供一个统一接口.Facade模式定义了一个高层接口,这个接口使得这子系统更容易使用. 2.UML 由于外观模式的结构图过于抽象,因此把它稍稍具体点. ...
- coffee 编译时, 用本地环境
export PATH="./node_modules/.bin:$PATH" 原文: How to use package installed locally in node_m ...
- [转]jQuery,javascript获得网页的高度和宽度
网页可见区域宽: document.body.clientWidth网页可见区域高: document.body.clientHeight网页可见区域宽: document.body.offsetWi ...
- php 钩子函数原理 解析
目前对钩子的理解:<转载:http://www.cnblogs.com/del/archive/2008/02/25/1080825.html> 譬如我们用鼠标在某个窗口上双击了一次, 或 ...
- Python中几种数据结构的整理,列表、字典、元组、集合
列表:shoplist = ['apple', 'mango', 'carrot', 'banana']字典:di = {'a':123,'b':'something'}集合:jihe = {'app ...