USACO Arithmetic Progressions

最新推荐文章于 2018-04-21 00:15:17 发布

neostar2008

最新推荐文章于 2018-04-21 00:15:17 发布

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文章标签： rest numbers output pair c++ 优化

本文链接：https://blog.youkuaiyun.com/neostar2008/article/details/7664989

本文探讨了如何优化寻找等差数列中所有项均可表示为两个平方数之和的问题，介绍了从暴力搜索到利用预计算数组加速查找的过程。

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判断一个数是否是完全平法数

对于给定一个整数N
定义一个整型数M
int M = sqrt(N);
然后判断
if(M*M == N)
则N是平方数;
else
N不是平方数;

第一遍代码结果和标准答案不一样，当时没有考虑到等差数列里的number拆解成两个平方数时，这两个平方根不能大于给定的M值——在处理（5,7）时我的程序给出了1,24的结果，实际上这个数列里有用到9^2 + 4^2，而9 大于 7 是不能使用的，在加入了平方根root小于M值的判定条件后，我的第二版程序在 test 6 时超时了……fuck，下面是第二版程序

/*
ID: wangxin12
PROG: ariprog
LANG: C++
*/

#include <iostream>
#include <fstream>
#include <vector>
#include <math.h>

using namespace std;
vector<int> ans;
int N, M;

bool isBiSquare(int number, int maxNum) ;

int main() {
	ifstream fin("ariprog.in");
	ofstream fout("ariprog.out");

	fin>>N;
	fin>>M;


	int maxNum = M * M * 2;
	int maxDiff = maxNum / (N - 1);

	int start = 0; //the first number in progression
	int diff = 1; //the difference between adjencent numbers
	int n = 0;

	for( diff = 1; diff <= maxDiff; diff++ ) {
		for(start = 0; start <= (maxNum - (N - 1) * diff); start++) {
			
			//for every unique start & diff pair, check if all numbers in this progression are bisquares
			bool flag = true;
			for(n = 0; n < N; n++) {
				//check
				if(!isBiSquare(start + n * diff, M)) {
					flag = false;
					break;
				}
			}
			//if check passed, output the answer
			if(flag) {
				//fout<<start<<" "<<diff<<endl;
				ans.push_back(start);
				ans.push_back(diff);
			}
		}
	}
	
	//cout<<isBiSquare(97,98);
	if(ans.empty()) {

		fout<<"NONE"<<endl;

	} else {
	
	for(int k = 0; k < ans.size(); k += 2) {
	
		fout<<ans[k]<<" "<<ans[k + 1]<<endl;

		}

	
	}



	fin.close();
	fout.close();

    return 0;
}




bool isBiSquare(int number, int M) {

	bool biFlag = false; 
	for(int i = 0; i * i <= number /2; i++) {
		int rest = number - i * i;

		int root =(int) sqrt((double)rest);
		if(root <= M && root * root == rest) {
			biFlag = true;
			break;
		}
	}

	return biFlag;
}

正在琢磨怎么优化……

在http://blog.youkuaiyun.com/vinsonou/article/details/7054669 上看到了优化方向——“上网看了一下别人的思路，原来是应该强搜双平方数，而不是强搜a b，双平方数最多有250*250*2 = 125000，可以接受，在双平方数中先强搜确定a，再根据a+nb<2*m*m确定b的强搜范围，对每个a b判断所有的序列是否属于双平方数。ACM虽然接触了一下，但仅仅限于选修课上。个人感觉每道题都不太会，看到题目只会强搜，深一点的算法题就不会了。但听了一位师兄说，只要坚持搞ACM一个月，一定会有感觉的。正在努力当中。。。”

于是下一版代码通过测试，主要修改了 bool isBiSquare()函数，将所有双平方数存入数组，再进行直接查找，比较方便。

/*
ID: wangxin12
PROG: ariprog
LANG: C++
*/

#include <iostream>
#include <fstream>
#include <vector>
#include <math.h>

using namespace std;

vector<int> ans;

int N, M;

bool biSquare[125000+10] = { false }; 

bool isBiSquare(int number) ;

int main() {
	ifstream fin("in.txt");
	ofstream fout("out.txt");

	fin>>N;
	fin>>M;
	
	//initial biSquare array to speed up
	for(int t1 = 0; t1 <= M; t1++) 
		for(int t2 = 0; t2 <= M; t2++) {
			int t = t1 * t1 + t2 * t2;
			if(biSquare[t] == false) biSquare[t] = true;
		}
	
	

	int maxNum = M * M * 2;
	int maxDiff = maxNum / (N - 1);

	int start = 0; //the first number in progression
	int diff = 1; //the difference between adjencent numbers
	int n = 0;

	for( diff = 1; diff <= maxDiff; diff++ ) {
		for(start = 0; start <= (maxNum - (N - 1) * diff); start++) {
			
			//for every unique start & diff pair, check if all numbers in this progression are bisquares
			bool flag = true;
			for(n = 0; n < N; n++) {
				//check
				if(!isBiSquare(start + n * diff)) {
					flag = false;
					break;
				}
			}
			//if check passed, output the answer
			if(flag) {
				ans.push_back(start);
				ans.push_back(diff);
				//fout<<start<<" "<<diff<<endl;
			}
		}
	}
	
	//cout<<isBiSquare(97,98);
	if(ans.empty()) {
		fout<<"NONE"<<endl;
	} else {
		for(int k = 0; k < ans.size(); k += 2) {
			fout<<ans[k]<<" "<<ans[k + 1]<<endl;
		}

	}



	fin.close();
	fout.close();

    return 0;
}

bool isBiSquare(int number) {
	return biSquare[number];
}

/*
bool isBiSquare(int number, int M) {
	bool biFlag = false; 

	for(int i = 0; i * i <= number /2; i++) {
		int rest = number - i * i;
		int root =(int) sqrt((double)rest);
		if(root <= M && root * root == rest) {
			biFlag = true;
			break;
		}
	}

	return biFlag;
}
*/