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…

题目大意: 给出一个真分数,把它分解成最少的埃及分数的和.同时给出了k个数,不能作为分母出现,要求解的最小的分数的分母尽量大. 分析: 迭代加深搜索,求埃及分数的基础上,加上禁用限制就可以了.具体可以参考一下紫书. #include<cstdio> #include<cstring> #include<algorithm> #include<set> using namespace std; typedef long long LL; LL ans[],v[…

[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…

题目传送门 /* x>=y, 1/x <= 1/y, 因此1/k - 1/y <= 1/y, 即y <= 2*k */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map> #include <…

UVA.10986 Fractions Again (经典暴力) 题意分析 同样只枚举1个,根据条件算出另外一个. 代码总览 #include <iostream> #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <sstream> #include <set> #include <map&g…

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…

直接暴力 没技巧 y应该从k+1开始循环,因为不然y-k<0的时候 你相当于(x*y) % (负数) 了. #include <iostream> using namespace std; ]; ]; int main() { int k,cnt; while(cin>>k) { cnt=; ;y<=*k;y++) { ) ) { X[cnt]=k*y/(y-k); Y[cnt]=y; cnt++; } } cout<<cnt<<endl; ;i…

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IDA* 就是iterative deepening(迭代深搜)+A*(启发式搜索) 启发式搜索就是设计估价函数进行的搜索(可以减很多枝哦~) 这题... 理论上可以回溯,但是解答树非常恐怖,深度没有明显上界,加数的选择理论上也是无限的. 我们可以从小到大枚举深度maxd, 设计估价函数,当扩展到第i层,前i个分数的和为c/d,第i的分数为1/e,接下来至少需要(a/b+c/d)/(1/e)个分数,如果超过maxd-i+1,那么直接回溯就好了.. #include<cstdio> #inclu…

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…