40if(i>=mll&&a2.length()>a1.length()){41arr4[i]=(((int)arr3[i]-48)+flag)%10;42flag=(((int)arr3[i]-48)+flag)/10;43}44if(i==ml-1&&flag==1){//确定最高位是否为进位得到的145arr4[i+1]=1;46}4748}495051if(arr4[ml]==0){//最高位为0则不输出最高位,52reverse(arr4...
分析A + B Problem II 题目 我有一个很简单的问题要问你。给定两个整数A和B,你的工作是计算A + B的和。 输入 输入的第一行包含整数T(1 < = T < = 20),这意味着测试用例的数量。然后T行遵循,每一行包含两个正整数,A和b。注意整数是非常大的,这意味着你不应该用32位整数来处理它们。您可以假设每...
A + B Problem II HDU - 1002 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 lines follow, each...
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, your job is to calculate the Sum of A +...
2)A + B = SUM 之后要换行 方法一: #include<stdio.h>#include<string.h>voidReverse(char*,int);intmain(void){chara[1003]={0};charb[1003]={0};charc[1003]={0};intn=0;intl=0;intla=0;intlb=0;intx=0;intnum=1;intzeroid[2]={0};scanf("%d",&n);for(inti=0;i<n;i++){ze...
include<stdio.h> include<stdlib.h> int main(){ int a,b;while(scanf("%d%d",&a,&b)!=EOF)printf("%d\n",a+b);system("pause");return 0;} 应该是正确的。。。
A+B Problem (II) : Input/Output Pratice Time Limit: 1 Sec Memory Limit: 2 MBSubmit: 571 Solved: 431[Submit][Status][Web Board]Description 计算a+b，0<=a,b<1000。Input 输入的第一行是一个整数N，后面有N对整数a和b，每对a和b占一行，a,b用空格分开。Output 每行输出一个a+...
int main(){ int t,i,len_str1,len_str2,len_max,k,num=1;char str1[1000];char str2[1000];scanf("%d",&t);getchar();while(t--){ int a[1000]={0},b[1000]={0},c[1001]={0};scanf("%s",str1);len_str1=strlen(str1);for( i=0;i<=len_str1-1;++i){ a[i...
Problem Description Ihave a very simple problemforyou.Giventwo integersAand B,your jobisto calculate theSumof 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 lines follow, ...
告诉你方法吧，用字符串输入，然后像做加法一样，对应位相加，然后看是否进位，进位的话下一位加一，用电脑本身的加法是不可能实现20多位的加法的，所以自己写个高精度加法，加几千位甚至上万位都是没问题的。问题补充：只能用字符串吗？回答：必须。