GTY's math problem Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 0    Accepted Submission(s): 0 Problem Description GTY is a GodBull who will get an Au in NOI . To have more time to learn alg…

题目链接:https://vjudge.net/problem/HDU-5171 GTY's birthday gift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1760    Accepted Submission(s): 685 Problem Description FFZ's birthday is coming. GTY…

A: HDU5170 这题让比较a^b与c^d的大小.1<=a,b,c,d<=1000. 显然这题没法直接做,要利用对数来求,但是在math库中有关的对数函数返回的都是浮点数,所以这又要涉及到eps问题. 其它就没有什么需要注意的了,我用的是log()函数,当然还可以用log10().....,原理不变. #include <iostream> #include <algorithm> #include <math.h> #include <map&g…

GTY's birthday gift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)  [Problem Description] FFZ's birthday is coming. GTY wants to give a gift to ZZF. He asked his gay friends what he should give to ZZF. One of them…

传送门 GTY's gay friends Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 264    Accepted Submission(s): 57 Problem Description GTY has n gay friends. To manage them conveniently, every morning he o…

#include <cstdio> #include <cstring> #include <vector> #include <algorithm> +; ],ans[MAX],t[MAX]; ],res[o<<|]); } ;     build(L,mid,o<<);     build(mid+,R,o<<|);     push_up(o); } ;     ;     ,ls,rs));     ,R,o<…

题目链接: hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5170 bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=567&pid=1001 题解: 先用java大数幂写的,t了 import java.util.*; import java.math.*; public class Main { public static void main(St…

数学题,a的b次方和c的d次方都很大,直接判断是做不出来的. 如果我们能找到一个函数F(x)是单调的,而F(X)的值又比较好算,那么可以通过比较F(X)的大小来判断自变量的大小. 令F(X)=log(X),a的b次方和c的d次方当做自变量. 那么接下来只要判断log(a的b次方)和log(c的d次方)的大小就可以判断a的b次方和c的d次方的大小了. 而log(a的b次方)=b*log(a),log(c的d次方)=d*log(c),很容易计算. 判断相等的时候注意一下精度问题. #include<…

题意: 给a,b,c,d. 比较a^b和c^d的大小 思路: 比较log(a^b)和log(c^d)的大小 代码: int a,b,c,d; int main(){ while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF){ double x1 = b*log((double)a); double x2 = d*log((double)c); if(fabs(x1-x2)<eps){ puts("=")…

Problem Description Jam has a math problem. He just learned factorization. He is trying to factorize ax^2+bx+cax​2​​+bx+c into the form of pqx^2+(qk+mp)x+km=(px+k)(qx+m)pqx​2​​+(qk+mp)x+km=(px+k)(qx+m). He could only solve the problem in which p,q,m,…