uva 10106】的更多相关文章

高精度乘法问题,WA了两次是因为没有考虑结果为0的情况.  Product  The Problem The problem is to multiply two integers X, Y. (0<=X,Y<10250) The Input The input will consist of a set of pairs of lines. Each line in pair contains one multiplyer. The Output For each input pair of…
package com.njupt.acm; import java.math.BigInteger; import java.util.Scanner; public class UVA_10106 { public static void main(String[] args) { Scanner scanner = new Scanner (System.in); while(scanner.hasNext()){ BigInteger a = scanner.nextBigInteger…
尝试一下java 的大数类 import java.util.*; import java.io.*; import java.math.BigInteger; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); while(in.hasNextBigInteger()) { BigInteger a = in.nextBigInteger(), b =…
 Product  The Problem The problem is to multiply two integers X, Y. (0<=X,Y<10250) The Input The input will consist of a set of pairs of lines. Each line in pair contains one multiplyer. The Output For each input pair of lines the output line should…
虽然是错的代码,但是还是想贴出来,最开始WA发现是没有考虑到乘积为0的情况,后来把a*0,0*a,a*0---0(若干个0),0--0(若干个0)*a都考虑进去了:可是还是WA,实在不懂先留在这儿.  Product  The Problem The problem is to multiply two integers X, Y. (0<=X,Y<10250) The Input The input will consist of a set of pairs of lines. Each…
Product The Problem The problem is to multiply two integers X, Y. (0<=X,Y<10250) The Input The input will consist of a set of pairs of lines. Each line in pair contains one multiplyer. The Output For each input pair of lines the output line should c…
打2017icpc沈阳站的时候遇到了大数的运算,发现java与c++比起来真的很赖皮,竟然还有大数运算的函数,为了以后打比赛更快的写出大数的算法并且保证不错,特意在此写一篇博客, 记录java的大数运算,也算是ACM java写法的入门: 学习博客:https://www.cnblogs.com/wkfvawl/p/9377441.html 进入到eclipse界面 第一步:file->new->java project->起名->finish 第二步:进入到刚才建的工程里,右键s…
白书一:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=64609#overview 注意UVA没有PE之类的,如果PE了显示WA. UVA401:Palindromes #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namesp…
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UVA - 10564 Paths through the Hourglass 题意: 要求从第一层走到最下面一层,只能往左下或右下走 问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径. f[i][j][k]从下往上到第i层第j个和为k的方案数 上下转移不一样,分开处理 没必要判断走出沙漏 打印方案倒着找下去行了,尽量往左走   沙茶的忘注释掉文件WA好多次   #include <iostream> #include <cstdio> #include <a…