杭电OJ 2054 A==B? A==B? 题目只要你判断数字A是否等于B,傻子都知道肯定不是int数据类型可以存下的,A可能有1000位,要用字符串类型存储,比较的时候注意0001、001、1都是一样的,即前导零不算 好吧,题目只说了number,没说是整数呀 分析: 比较前,先去掉前面多余的0,再去小数点后面多余的0(首先要保证有小...
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 line consists of two positive integers, A and B. Notice that the integers ...
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 lines follow, each line consists of ...
若记b[j]=max(a[i]+a[i+1]+..+a[j]),其中1<=i<=j,并且i<=j<=n。则所求的最大子段和为max b[j],1<=j<=n。 由b[j]的定义可易知,当b[j-1]>0时b[j]=b[j-1]+a[j],否则b[j]=a[j]。故b[j]的递推方程为: b[j]=max(b[j-1]+a[j],a[j]),1<=j<=n。 笔者的...
HDOJ(HDU) 1877 又一版 A+B(进制、、) Problem Description 输入两个不超过整型定义的非负10进制整数A和B(<=231-1),输出A+B的m (1 < m <10)进制数。 Input 输入格式:测试输入包含若干测试用例。每个测试用例占一行,给出m和A,B的值。 当m为0时输入结束。 Output 输出格式:每个测试用例的输出占...
杭电oj: 1000 A + B Problem Java实现 A + B Problem import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); while(sc.hasNextInt()) { int a = sc.nextInt();...
each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000. Output
hdu 2057 A + B again 题解: 乍一看以为是个模拟题,但是仔细看了一下题意,才发现最大数字是15位,没有超过 __int64 能表达的最大数。__int64 最大能表示16位十六进制数。所以直接进行加减就行了,需要注意的是输入输出的格式以及十六进制表达负数的时候显示的是补码。 代码如下: 转载于:https://www....
1002 A + B 我的解法通过.cpp 我的代码通过方法 1。清零【数组要放for(在下面】 2。回车【最末一行只输出一个回车】 3。布尔类型【#include <stdbool.h>//C语言专用】 #include <stdio.h> #include <string.h> //#include <stdbool.h>//C语言专用 ...
#include<stdio.h>#include<string.h>voidSum(char a[],char b[],int c[],int lena,int lenb);intmain(void){char a[1000]={0},b[1000]={0};int c[1001]={0};int number,i,j,lena,lenb,max;scanf("%d",&number);for(i=0;i<number;i++){/***初始化***/char a[1000]={0};char...