题意:问方程X^Z + Y^Z + XYZ = K (X<Y,Z>1)有多少个正整数解 (K<2^31) 解法:看K不大,而且不难看出 Z<=30, X<=sqrt(K), 可以枚举X和Z,然后二分找Y,这样的话不把pow函数用数组存起来的话好像会T,可以先预处理出1~47000的2~30次幂,这样就不会T了. 但是还可以简化,当Z=2时,X^2+Y^2+2XY = (X+Y)^2 = K, 可以特判下Z= 2的情况,即判断K是否为平方数,然后Z就可以从3开始了,这样的话X^…
A very hard mathematic problem Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4282 Description Haoren is very good at solving mathematic problems. Today he is working a problem like this: Find three positive in…
由于k的范围是0-2^31,而且x,y,z都是正整数,由题易知道2<=z<31,1<=x<y;所以直接枚举就好了!!! #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<stdlib.h> #include<cstring> #include<v…
pid=4972" target="_blank" style="">题目链接:hdu 4972 A simple dynamic programming problem 题目大意:两支球队进行篮球比赛,每进一次球后更新比分牌,比分牌的计数方法是记录两队比分差的绝对值,每次进球的分可能是1,2,3分. 给定比赛中的计分情况.问说最后比分有多少种情况. 解题思路:分类讨论: 相邻计分为1-2或者2-1的时候,会相应有两种的的分情况 相邻计分之差大于3或…
HDU 3081 Marriage Match II (网络流,最大流,二分,并查集) Description Presumably, you all have known the question of stable marriage match. A girl will choose a boy; it is similar as the game of playing house we used to play when we are kids. What a happy time as…
The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .…
Yukari's Birthday  HDU4430 就是枚举+二分: 注意处理怎样判断溢出...(因为题目只要10^12) 先前还以为要用到快速幂和等比数列的快速求和(但肯定会超__int64) 而且这样判断会超时的... 还有题目中的And it's optional to place at most one candle at the center of the cake. (中间的蜡烛可有可无) 还有观察数据就知道:因为n最大10^12,r最多枚举到40,然后二分k的结果,看是否有符合条…
1514: Packs Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 61  Solved: 4[Submit][Status][Web Board] Description Give you n packs, each of it has a value v and a weight w. Now you should find some packs, and the total of these value is max, total of…
HDU 1588 Gauss Fibonacci(矩阵高速幂+二分等比序列求和) ACM 题目地址:HDU 1588 Gauss Fibonacci 题意:  g(i)=k*i+b;i为变量.  给出k,b,n,M,问( f(g(0)) + f(g(1)) + ... + f(g(n)) ) % M的值. 分析:  把斐波那契的矩阵带进去,会发现这个是个等比序列. 推倒: S(g(i)) = F(b) + F(b+k) + F(b+2k) + .... + F(b+nk) // 设 A = {1…
http://acm.hdu.edu.cn/showproblem.php?pid=4282 对于方程X^Z + Y^Z + XYZ = K,已知K求此方程解的个数,其中要求X<Y,Z>1,而K的范围是0到2^31. 首先我们来分析Z的范围:由于X,Y为正整数,X < Y,则1 < X < Y, =====> Y >= 2 => X^Z + Y^Z + XYZ > Y^Z => 2^Z <= Y^Z < 2^31 所以得到2 <…