UVA12558 Egyptian Fractions (HARD version) 题解 迭代加深搜索,适用于无上界的搜索.每次在一个限定范围中搜索,如果无解再进一步扩大查找范围. 本题中没有分数个数和分母的上限,只用爆搜绝对TLE.故只能用迭代加深搜索. #include<cstdio> #include<cstring> #include<set> using namespace std; typedef long long ll; int num,T,t,k;…
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 迭代加深搜索. 枚举最大量maxdep 在dfs里面传剩余的要凑的分子.分母 以及上一次枚举的值是多少. 然后找到最小的k,满足1/k<=分子/分母 然后从max(k,last+1)开始枚举. ->剪枝就是剩余的全都用这个最大的分数.如果都不行就肯定不行了. 二分找这个k. 不能用的数字就直接跳过就行. [代码] /* 1.Shoud it use long long ? 2.Have you ever test severa…
Problem UVA12558-Efyptian Fractions(HARD version) Accept:187 Submit:3183 Time Limit: 3000 mSec Problem Description Given a fraction a/b, write it as a sum of different Egyptian fraction. For example, 2/3 = 1/2 + 1/6. Thereisonerestrictionthough: th…
题目大意:给一个分数,对其进行化简.因为分子.分母最大为1030,所以用要用大数. import java.io.*; import java.util.*; import java.math.*; class Main { public static void main(String[] args) { Scanner cin = new Scanner(System.in); int T = cin.nextInt(); while (T-- > 0) { BigInteger a = ci…
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] x>=y => \(\frac{1}{x}<=\frac{1}{y}\) => \(\frac{1}{x}=\frac{1}{k}-\frac{1}{y}\) 结合两个式子可以得到 y<=2*k 则枚举y,然后根据式子得到x,判断合法性就ok [代码] /* 1.Shoud it use long long ? 2.Have you ever test several sample(at least ther…
Problem EEg[y]ptian Fractions (HARD version)Given a fraction a/b, write it as a sum of different Egyptian fraction. Forexample, 2/3=1/2+1/6.There is one restriction though: there are k restricted integers that should notbe used as a denominator. For…
10976 Fractions Again It is easy to see that for every fraction in the form 1 k (k > 0), we can always find two positive integers x and y, x ≥ y, such that: 1 k = 1 x + 1 y Now our question is: can you write a program that counts how many such pairs…
It is easy to see that for every fraction in the form (k > 0), we can always find two positive integers x and y,x ≥ y, such that: . Now our question is: can you write a program that counts how many such pairs of x and y there are for any givenk? Inp…
基本思路就是Bfs: 本题的一个关键就是如何判段状态重复. 1.如果将状态用一个int型数组表示,即假设为int state[17],state[0]代表机器人的位置,从1到M从小到大表示障碍物的位置.那么如果直接用STL中的set是会超时的,但如果自己建立一个hash方法,像这样: int getKey(State& s) { long long v = 0; for(int i=0; i<=M; ++i ) { v = v * 10 + s[i]; } return v % hashSi…
题目链接:https://vjudge.net/problem/UVA-10976 It is easy to see that for every fraction in the form 1k(k > 0), we can always find two positive integers x and y, x ≥ y, such that: 1/k=1/x+1/y Now our question is: can you write a program that counts how ma…
[题意]:给你一个数k,求所有使得1/k = 1/x + 1/y成立的x≥y的整数对. [分析]:枚举所有在区间[k+1, 2k]上的 y 即可,当 1/k - 1/y 的结果分子为1即为一组解. [代码]: #include<bits/stdc++.h> using namespace std; int x[10005]; int y[10005]; #define LL long long int main() { int k,c; while(~scanf("%d",…
It is easy to see that for every fraction in the form 1k(k > 0), we can always find two positive integersx and y, x ≥ y, such that:1k=1x+1yNow our question is: can you write a program that counts how many such pairs of x and y thereare for any given…
题目链接: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=995 Problem D: The Necklace My little sister had a beautiful necklace made of colorful beads. Two successive beads in the necklace shared a c…
参考:http://www.cnblogs.com/xiaobaibuhei/p/3301110.html 算法学到很弱,连这么简单个问题都难到我了.但我偏不信这个邪,终于做出来了.不过,是参照别人的,是 xiaobaibuhei 到博客让我找到到感觉,不过我只看了一遍他到代码,后面都是自己写的,虽然写的很像... Write a program that finds and displays all pairs of 5-digit numbers that between them use…
题目地址: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=258 http://poj.org/problem?id=1138 题目描写叙述: Ships Probably everyone who ever attended school knows the game where two opposing players place…
