大数加法(A + B Problem Plus)问题

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大数加法(A + B Problem Plus)问题

解题思路两个⼤数可以⽤数组来逐位保存,然后在数组中逐位进⾏相加,再判断该位相加后是否需要进位.为了⽅便计算,可以把数字的低位放到数组的前面,高位放在后面. 样例输入 3 18 22 56 744 53234673473 7544645548767 样例输出 Case 1: 18 + 22 = 40 Case 2: 56 + 744 = 800 Case 3: 53234673473 + 7544645548767 = 7597880222240 代码实现 #include<stdio.h>…

A + B Problem II（大数加法）

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): 261608 Accepted Submission(s): 50625 Problem Description I have a very simple p…

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…

大数加法~HDU 1002 A + B Problem II

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1002 题意: 数学题,A+B; 思路,这个数非常大,普通加法一定会超时,所以用大数加法.大数加法的基本思路是模拟我们做加法的时候的进位思想,从最低位开始模拟, 在我的代码里 v -- 进位, tmp -- 当前位2个数的和,k -- 答案数组的下标. 其中, tmp/10 可以得到要进的位数, 同理,tmp%10可以得到个位,就是答案数组对应的位下面是AC代码: #include <iostre…

hdu 1002 A + B Problem II【大数加法】

题目链接>>>>>> 题目大意:手动模拟大数加法,从而进行两个大数的加法运算 #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; #define MAXN 3000 int a[MAXN], b[MAXN]; int main() { int t; scanf(; getchar(); int array[MAXN]; whi…

玲珑杯1007-A 八进制大数加法（实现逻辑陷阱与题目套路）

题目连接:http://www.ifrog.cc/acm/problem/1056 DESCRIPTION Two octal number integers a, b are given, and you need calculate the result a - b in octal notation.If the result is negative, you should use the negative sign instead of complement notation. INPU…