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…

原题代号:HDU 1002 原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002 原题描述: Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.   Input The first line of the input contains an i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002 题目意思:就是大整数加法. 两年几前做的,纯粹是整理下来的. #include <stdio.h> #include <string.h> #define max 1010 char a[max], b[max]; int main() { int i, k, s, c, T, len1, len2; scanf("%d", &T); ; k <…

A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 230395    Accepted Submission(s): 44208 Problem Description I have a very simple problem for you. Given two integers A and B, you…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1002 复习一下大数 模板: #include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #include <assert.h> #include <ctype.h> #include <map> #include <string&g…

http://acm.hdu.edu.cn/showproblem.php?pid=1002 PE   由于最后一个CASE不须要输出空行 import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String args[]){ Scanner in = new Scanner(System.in); int ncase; ncase = in.nextIn…

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…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002 Problem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. InputThe first line of the input contains an integer T(1<=T<=20) whic…

昨天做的那题其实和大整数相加类似.记得当初我大一寒假就卡在这1002题上,结果之后就再也没写题... 到今天终于把大整数相加写了一遍. 不过写的很繁琐,抽时间改进一下写简洁一点. #include<queue> #include<math.h> #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std;…

ps:wa了好多次,然后才发现是输入的时候%s和%s要隔开一个空格,我想当然了... 代码: #include "stdio.h" #include "string.h" ]; ]; ]; ]; int cal(); int main(){ ; scanf("%d",&T); getchar(); while(T--){ scanf("%s %s",&a,&b); memset(c,'\0',sizeo…

A + B Problem II   Time Limit: 1000MS      Memory Limit: 65536K Total Submissions: 16104    Accepted: 4547 Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. Input The first line o…

#include <stdio.h> #include <string.h> #define MAX 1009 int main() { },b[MAX]={}; ,z=,r=,k=; char *p=NULL; scanf("%d",&n); <n&&n< ){ ;i<n;i++){ j=; z=; r=; scanf("%s",a); scanf("%s",b); printf…

package Main; //import java.io.InputStream; import java.math.BigDecimal; import java.util.Scanner; public class Main { public static void main(String[] args) { int _case; BigDecimal a,b; a=new BigDecimal("0"); b=new BigDecimal("0"); Sc…

A + B Problem II Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.    Input The first line of the inpu…

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…

A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 407964    Accepted Submission(s): 79058 Problem Description I have a very simple problem for you. Given two integers A and B, you…

A + B ProblemII Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 317773    Accepted Submission(s):61748 Problem Description I have a verysimple problem for you. Given two integers A and B, your j…

题目链接 Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.   Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T li…

A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 435932    Accepted Submission(s): 84825 Problem Description I have a very simple problem for you. Given two integers A and B, you…

题意:就是求a+b (a,b都不超过1000位) 思路:用数组存储 第一道大数的题目,虽然很水,纪念一下! 代码: #include<cstdio> #include<cstring> int main(){ int t,lena,lenb,i,j,sum,testcase=0; char a[1024],b[1024]; char c[1024];//存 a+b int co;//进位 scanf("%d",&t); int tt=t; while(t…

#include <iostream> #include <cstring> #include <string> #include <cstdio> #include <algorithm> using namespace std; string bigmun(string a,string b){     string c;     reverse(a.begin(),a.end());     reverse(b.begin(),b.end(…

代码: #include<cstdio> #include<cstring> #define Min(a,b) ((a)<(b)?(a):(b)) using namespace std; char s1[10000],s2[10000],s3[10000]; int main() { int t; scanf("%d",&t); int tt=0; while(t--) { ++tt; if(tt!=1) printf("\n"…

A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 318065    Accepted Submission(s): 61825 Problem Description I have a very simple problem for you. Given two integers A and B, you…

链接:传送门 题意:A + B 高精度,板子题 /************************************************************************* > File Name: hdu1002.cpp > Author: WArobot > Blog: http://www.cnblogs.com/WArobot/ > Created Time: 2017年05月16日 星期二 20时08分28秒 *******************…

Constructing Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20927    Accepted Submission(s): 8023 Problem Description There are N villages, which are numbered from 1 to N, and you should…

非常简单的大数加法,因为不会Java只能手写大数加法了;博客存一下以后回来看看 #include<bits/stdc++.h> using namespace std; +; char A[maxn],B[maxn],ans[maxn]; void flip(char str[]) { ,endstr=strlen(str)-; for(;startstr<endstr;startstr++,endstr--) swap(str[startstr],str[endstr]); } voi…

HDU 1000 A + B Problem  I/O HDU 1001 Sum Problem  数学 HDU 1002 A + B Problem II  高精度加法 HDU 1003 Maxsum  贪心 HDU 1004 Let the Balloon Rise  字典树,map HDU 1005 Number Sequence  求数列循环节 HDU 1007 Quoit Design  最近点对 HDU 1008 Elevator  模拟 HDU 1010 Tempter of th…

前几天这篇博客写了太多废话! 删啦~~. 本篇博客只是为chd A协的全嫩小鲜肉入门C语言的预科, 如果你在此处学习C语言, 不幸走火入魔, 小弱概不负责. //请直接随便找个C语言编译器,抄一下下面的程序 //如果不会用编译器,或根本不知什么是编译器 //请自行默默百度, 下载, 安装,学习使用 //如果你懒得不想去百度, 请出门遇红灯直行! #include<stdio.h> int main() { printf("What fuck are you doing?")…

http://codeforces.com/contest/747/problem/F cf #387 div2 problem f 非常好的一道题.看完题,然后就不知道怎么做,感觉是dp,但是不知道怎么枚举.还有就是一般求第k小的思路一般是什么?对这类题目理解的不是很好. 这道题,跟上一篇的题解里面写的hdu 1002 递增数的题目的解法差不多,但是底层的求解需要花一些时间去推敲! 不会做,就看官方题解,看不懂说的什么.就从下面的讨论找大神的题解. I can describe my solu…