【hdoj 1002 A + B Problem II【大数加法】】的更多相关文章

hdoj 1002 A + B Problem II【大数加法】

A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 260585    Accepted Submission(s): 50389 Problem Description I have a very simple problem for you. Given two integers A and B, you…

HDU 1002 A + B Problem II(高精度加法(C++/Java))

A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 347161    Accepted Submission(s): 67385 Problem Description I have a very simple problem for you. Given two integers A and B, you…

hdoj 1002 A + B Problem II

A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 242959    Accepted Submission(s): 46863 Problem Description I have a very simple problem for you. Given two integers A and B, you…

A + B Problem II 大数加法

题目描述: Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should n…

hdoj 1002 A + B Problem II 高精度 java

A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 241933    Accepted Submission(s): 46646 Problem Description I have a very simple problem for you. Given two integers A and B, you…

hdu1002 A + B Problem II[大数加法]

目录 题目地址 题干 代码和解释 参考 题目地址 hdu1002 题干 代码和解释 由题意这是一个涉及到大数的加法问题.去看了一眼大数加法的方法感觉头很大,然后突然发现Java可以流氓解决大数问题,毅然为之. import java.util.Scanner; import java.math.BigInteger; public class Main{ public static void main(String args[]){ BigInteger a,b; int T; int n=1;…

A + B Problem II(大数加法）

一直格式错误,不想改了,没A #include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> using namespace std; int main() { ; scanf("%d",&T); ],b[]; ],tb[]; getchar(); while(T--) { K++; scanf("%s%s",a,b)…

抓起根本（二）（hdu 4554 叛逆的小明 hdu 1002 A + B Problem II,数字的转化（反转），大数的加法......）

数字的反转: 就是将数字倒着存下来而已.(*^__^*) 嘻嘻…… 大致思路:将数字一位一位取出来,存在一个数组里面,然后再将其变成数字,输出. 详见代码. while (a) //将每位数字取出来,取完为止 { num1[i]=a%; //将每一个各位取出存在数组里面,实现了将数字反转 i++; //数组的变化 a/=; } 趁热打铁 例题:hdu 4554 叛逆的小明 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4554 叛逆的小明 Time…

hdu1002 A + B Problem II(大数题)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002 A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 230247    Accepted Submission(s): 44185 Problem Description I have a very sim…

Problem : 1002 ( A + B Problem II )

经验总结:一定要注意输出的格式,字符的空格,空行,一定要观察清楚.如本题的最后一个输出结果后面没有空行.最后代码实现的时候需要判断一下,代码如下 !=n) cout<<endl; Problem : 1002 ( A + B Problem II )     Judge Status : Accepted RunId : 10370349    Language : C++    Author : mnmlist Code Render Status : Rendered By HDOJ C+…