Pat 1001. A+B format

此题为两个数求和后进行格式化输出，因为没有超出Integer的表示范围，可以直接整数求和。

思路如下：

import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;public class Main {  public static void main(String[] args) throws IOException {    // TODO Auto-generated method stub    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));    String str = br.readLine();    String[] strs = str.split(" ");    int a = Integer.parseInt(strs[0]);    int b = Integer.parseInt(strs[1]);    int sum = a+b;    if(sum<0){      System.out.PRint("-");      sum=-sum;    }	//格式化输出 if(sum>=1000000){      System.out.printf("%d,%03d,%03d/n",sum/1000000,(sum/1000)%1000,sum%1000);    }else if(sum>=1000){      System.out.printf("%d,%03d/n",sum/1000,sum%1000);    }else{      System.out.printf("%d/n",sum);    }    return;    }}如果题目变一下，变为两个大数相加后格式化输出，这时就会超出Integer的表示范围。所以需要转化为字符串相加减。思路如下：
import java.io.BufferedReader;import java.io.InputStreamReader;public class Main {	public static void main(String[] args) throws Exception {		// TODO Auto-generated method stub		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));		String str = br.readLine();		String[] strs = str.split(" ");		String str1 = strs[0];		String str2 = strs[1];		System.out.println(minusBigInteger(str1, str2));		System.out.println(addBigInteger(str1, str2));	}		private static String addBigInteger(String str1,String str2){		int i = str1.length()-1;		int j = str2.length()-1;		int takeOver =0;		int sum;		String result="";		while(i>=0&&j>=0){			sum = str1.charAt(i)-'0'+str2.charAt(j)-'0'+takeOver;			if(sum>9){				sum = sum%10;				takeOver =1;			}else{				takeOver =0;			}			result+=(char)(sum+'0');			i--;			j--;		}		while(i>=0){			if(takeOver>0){				sum = str1.charAt(i)-'0'+takeOver;				if(sum>9){					sum = sum%10;					takeOver =1;				}else{					takeOver =0;				}				result+=(char)(sum+'0');			}else{				result+=new StringBuffer(str1.substring(0, i+1)).reverse();				break;			}			i--;		}		while(j>=0){			if(takeOver>0){				sum = str2.charAt(j)-'0'+takeOver;				if(sum>9){					sum = sum%10;					takeOver =1;				}else{					takeOver =0;				}				result+=(char)(sum+'0');			}else{				result+=new StringBuffer(str2.substring(0, j+1)).reverse();								break;			}			j--;		}		if(takeOver>0){			result+=takeOver;		}		return new StringBuffer(result).reverse().toString();	}	private static String minusBigInteger(String str1,String str2){		//始终为str1-str2		int i=str1.length()-1;		int j=str2.length()-1;		String result ="";		int takeOver=0;		int sum;		while(i>=0&&j>=0){			sum=(str1.charAt(i)-'0')-(str2.charAt(j)-'0')-takeOver;			if(sum<0){				sum+=10;				takeOver=1;			}else{				takeOver=0;			}			result+=(char)(sum+'0');			i--;			j--;		}		while(i>=0){			if(takeOver>0){				sum=str1.charAt(i)-'0'-takeOver;				if(sum<0){					sum+=10;					takeOver=1;				}else{					takeOver=0;				}				result+=(char)(sum+'0');			}else{				result+=new StringBuffer(str1.substring(0, i)).reverse();				break;			}			i--;		}		int k =result.length()-1;		while(result.charAt(k)=='0'){			k--;		}		result=result.substring(0, k+1);		return new StringBuffer(result).reverse().toString();	}	}运行结果如下图