A + B Problem II
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 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
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line is the an equation “A + B = Sum”, Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
如果用C语言或者C++都需要用数组去模拟竖式，特别的恶心，代码一撸就是几十行，挺恶心的。
#include <stdio.h>
#include <string.h>
char a[1001];
char b[1001];
char c[1001];
void swap(char target[]) {
int i;
int j;
char temp;
for(i = 0,j = strlen(target) - 1;i <= j;i++,j--) {
temp = target[i];
target[i] = target[j];
target[j] = temp;
}
}
void add(char a[],char b[]) {
int i;
for(i = 0;i < strlen(a) && i < strlen(b);i++) {
c[i] += a[i] + b[i] - '0';
if(c[i] - '0' >= 10){
c[i] = c[i] - 10;
c[i+1] = 1;
}
}
//先将对其的进行运算
if(strlen(a) == strlen(b)) {
if(c[i] == 1)
c[i]='1';
}
//长度相等，看是否进一位
if(strlen(a) > strlen(b)){
if(c[i] == 1) {//进位的话就得每位进行累加
for(;i < strlen(a);i++){
c[i] += a[i];
if(c[i] - '0' >= 10) {
c[i] = c[i] - 10;
c[i+1] = 1;
}
}
if(c[i-1] == '0')//最后一位进位
c[i] = '1';
}
else { //不进位的话就直接把数组接在后面
for(;i < strlen(a);i++)
c[i] = a[i];
}
}
if(strlen(b) > strlen(a)){
if(c[i]==1){
for(;i < strlen(b);i++){
c[i] += b[i];
if(c[i] - '0' >= 10){
c[i] = c[i] - 10;
c[i+1] = 1;
}
}
if(c[i] == 1)
c[i] = '1';
} else {
for(;i < strlen(b);i++)
c[i] = b[i];
}
}
}
int main(void) {
scanf("%s",a);
scanf("%s",b);
swap(a);
swap(b);
add(a,b);
swap(c);
printf("%s\n",c);
return 0;
} 后来看一个同学比赛的比赛时一道相似题的题解http://blog.csdn.net/hepangda/article/details/73431191
才知道Java的BigInteger做这种题特别轻松，而且不需要担心溢出
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int T;
int count = 1;
Scanner cin = new Scanner(System.in);
T = cin.nextInt();
while(T-- > 0) {
String s1 = cin.next();
String s2 = cin.next();
BigInteger a = new BigInteger(s1);
BigInteger b = new BigInteger(s2);
b = a.add(b);
System.out.println("Case " + count + ":");
System.out.println(s1 + " + " + s2 + " = " + b);
count++;
if(T > 0) {
System.out.println();
}
}
cin.close();
}
}对于大数加减乘法，无非就是调用bigInteger的add(),subtract(),multiply方法,但不能直接进行加减乘除，只能通过调用方法来完成
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String []args){
Scanner cin = new Scanner(System.in);
while(cin.hasNext()) {
BigInteger a = cin.nextBigInteger();
BigInteger b = cin.nextBigInteger();
BigInteger c1 = a.add(b); //大数加法
System.out.println("加的结果为 "+c1);
BigInteger c2 = a.subtract(b); //大数减法
System.out.println("减的结果为 "+c2);
BigInteger c3 = a.multiply(b); //大数乘法
System.out.println(c3);
BigInteger c4 = a.divide(b); //大数除法
System.out.println("除的结果为 "+c4);
BigInteger c5 = a.mod(b); //大数模
BigInteger c51 = a.remainder(b);
System.out.println("模的结果为 "+c51);
BigInteger c6 = a.max(b);//取最大
System.out.println("最大为 "+c6);
BigInteger c7 = a.min(b); //取最小
System.out.println("最小为 "+c7);
if(a.equals(b)) { //判断是否相等
System.out.println("相等");
}else {
System.out.println("不相等");
}
BigInteger c8 = a.gcd(b); //求最大公约数
System.out.println("最大公约数为 "+c8);
}
cin.close();
}
}