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  • Codeforces Gym101502 H.Eyad and Math-换底公式

    H. Eyad and Math
     
    time limit per test
    2.0 s
    memory limit per test
    256 MB
    input
    standard input
    output
    standard output

    Eyad was given a simple math problem, but since he is very bad at math he asked you to help him.

    Given 4 numbers, abc, and d. Your task is to find whether ab is less than cd or not.

    It is guaranteed that the two numbers above are never equal for the given input.

    Input

    The first line contains an integer T (1 ≤ T ≤ 105), where T is the number of test cases.

    Then T lines follow, each line contains four integers abc, and d (1 ≤ a, b, c, d ≤ 109).

    Output

    For each test case, print a single line containing "<" (without quotes), if ab is less than cd. Otherwise, print ">" (without quotes).

    Example
    input
    2
    9 2 5 3
    3 4 4 3
    output
    <
    >

     这个题,emnnn,眼瞎,这么大的数,快速幂也跑爆啊。。。中学时代的换底公式,6的飞起。。。

    a的b次方假设为x,求对数为logax=b,换成以10为底的,就是lnx为b*lna,同理c的d次方。水题水题。。。

    代码:

     1 //H. Eyad and Math-对数,换底公式-快速幂会爆啊啊啊啊啊啊
     2 #include<iostream>
     3 #include<cstring>
     4 #include<cstdio>
     5 #include<cmath>
     6 #include<algorithm>
     7 using namespace std;
     8 int main(){
     9     int t;
    10     while(~scanf("%d",&t)){
    11     int a,b,c,d;
    12     while(t--){
    13         scanf("%d%d%d%d",&a,&b,&c,&d);
    14         double ans1=b*log10(a);
    15         double ans2=d*log10(c);
    16         if(ans1<ans2)printf("<
    ");
    17         else printf(">
    ");
    18     }
    19     }
    20     return 0;
    21 }
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  • 原文地址:https://www.cnblogs.com/ZERO-/p/9703304.html
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