A1058 A+B in Hogwarts

阿新 • • 發佈：2018-12-02

不解釋

#include <stdio.h>

int main(int argc, char const *argv[])
{
	int G1,S1,K1,G2,S2,K2;
	scanf("%d.%d.%d %d.%d.%d",&G1,&S1,&K1,&G2,&S2,&K2);
	int carry = 0;
	int g,s,k;
	k = (K1+K2)%29;
	carry = (K1+K2)/29;
	s = (S1+S2+carry)%17;
	carry = (S1+S2+carry)/17;
	g = G1+ 
G2+carry;
	printf("%d.%d.%d",g,s,k);
	return 0;
}

A1058 A+B in Hogwarts

不解釋 #include <stdio.h> int main(int argc, char const *argv[]) { int G1,S1,K1,G2,S2,K2; scanf("%d.%d.%d %d.%d.%d",&G1,&S1,&K

PAT1058:A+B in Hogwarts

name key col bst while size title plain 處理 1058. A+B in Hogwarts (20) 時間限制 50 ms 內存限制 65536 kB 代碼長度限制 16000 B 判題程序 Standard 作者 C

1058. A+B in Hogwarts (20)

you turn fan explained names world printf blog tput If you are a fan of Harry Potter, you would know the world of magic has its own curre

PAT-ADVANCED1058——A+B in Hogwarts

我的PAT-ADVANCED程式碼倉：https://github.com/617076674/PAT-ADVANCED 原題連結：https://pintia.cn/problem-sets/994805342720868352/problems/994805416519647232

1058 A+B in Hogwarts

題目大意： 1 Galleon=17 Sickle，1 Sickle=29 Knut。計算A+B。解題思路：模擬。程式碼如下： #include<iostream> #include<cstdio> #include<fstrea

PAT 甲級 1058 A+B in Hogwarts [Java實現]

1. 題意也是進位制轉換的題目兩個數字相加，然後第二項是17進位制，第三項是29進位制 2. 思路水題。注意最高位沒有進位限制，一開始被這個坑了。 3. 程式碼 package adv1058; import java.io.BufferedReader;

PAT 1058 A+B in Hogwarts （20 分）

1058 A+B in Hogwarts （20 分） If you are a fan of Harry Potter, you would know the world of magic has its own currency system – as Ha

【PAT-A】1058. A+B in Hogwarts 寫題記錄

#include <cstdio> struct man{ int G; //[0,10^7] int S; //[0,17) int K; //[0,29) }A,B; int main(){ int count=0; scanf("%d.%d.%d",&A.G,&a

1058 A+B in Hogwarts （20 分）

If you are a fan of Harry Potter, you would know the world of magic has its own currency system -- as Hagrid explained it to Harry, "Seventeen silver

1058 A+B in Hogwarts （20 分）混合進位制

題目 If you are a fan of Harry Potter, you would know the world of magic has its own currency system – as Hagrid explained it to Harry, “Sevente

PAT (Advanced Level) Practice 1058 A+B in Hogwarts （20 分）（C++）（甲級）

1058 A+B in Hogwarts （20 分） If you are a fan of Harry Potter, you would know the world of magic has its own currency system – as Hagrid explai

PAT甲級水題 A+B in Hogwarts（java string中分隔符的用法）

說明 c中 wid html get mage 轉義字符 split方法 bsp A+B in Hogwarts 在java.lang包中有String.split()方法,返回是一個數組我在應用中用到一些,給大家總結一下,僅供大家參考: 1、如果用“.”作為分

1058. A+B in Hogwarts (20):有疑惑-第三個測試點錯誤

也考慮到資料的區間越界問題了··· #include<iostream> #include<cstring> #include<map> #include<string> #include<cmath> #incl

SPOJ4491. Primes in GCD Table（gcd(a,b)=d素數，（1<=a<=n,1<=b<=m））加強版

function ted solid result writing set silver %d ron SPOJ4491. Primes in GCD Table Problem code: PGCD

python裡面的幾個用法，not in，c if a else b，[fun(a) for a in [...]] , a,b=b,a,'內容'.join([string array])

1.not in >>> a=2 >>> a not in [2,3,4] False >>> a in [2,3,4] 2. c if a else b #這裡注意，一定要有b,而且b不

Python中的 A for B in C 用法

一個例子 Y = [ [ int(x1+x2 < 1) ] for (x1, x2) in X ] 對X中的每一組元素(x1, x2)遍歷一遍，當滿足(x1+x2 < 1)時，就把這個布林值

Multiple Responses in A/B Testing

Multiple Responses in A/B Testingtl;drOur data scientists use previous experiments to build a prior on the relationships of treatment effects between sever

Calculate Rotation Matrix to align Vector A to Vector B in 3d?

Octave/Matlab Implementation The basic implementation is very simple. You could improve it by factoring out the common expressions of dot(A,

python之列表推導式：c for b in a的程式碼形式理解

直接上程式碼可能更直觀點： y= [x*x for x in range(10)] print(y) 輸出的是一個列表： [0, 1, 4, 9, 16, 25, 36, 49, 64, 81] 這段程式碼類似的就是： for x in range(10):

使用Java語言，使用System.in輸入以下各個變數的值，並按照公式計算得出結果。公式為： 4/3(r+34)-9(a+bc)+(3+d(2+a))/(a+b*d)

public class Demo2 { /** * @param args */ /* * 根據題目要求，編寫程式完成相應要求：使用Java語言，使

A1058 A+B in Hogwarts

相關推薦