HDOJ-1002大数相加

发布时间：2020-12-14 02:37:15 所属栏目：大数据来源：网络整理

导读：参考这篇：点击打开链接 ? 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 numb

参考这篇：点击打开链接

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 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 = 2222222222222221110

#include <iostream>
#include<cstdio>
#include<string>
#include<string.h>
void add(char a[],char b[],char c[]);
int main(int argc,char *argv[]) {	
	int x;
	scanf("%d",&x);
	for(int i=1;i<=x;i++){	
			char a[1005],b[1005],c[1005];
			a[0]=b[0]='0';//数组首位置0，方便最高位进位 
			scanf("%s",&a[1]);
			scanf("%s",&b[1]);
			printf("Case %d:n",i);	
			add(a,b,c);
			printf("%s + %s = ",&a[1],&b[1]);
			if(c[0]!='0')printf("%c",c[0]);
			printf(i==x?"%sn":"%snn",&c[1]);
	}
	return 0;
	
}
void add(char a[],char c[]){
	int lenA,lenB,lenC;
	lenA=strlen(a);
	lenB=strlen(b);
	lenC=(lenA>lenB)?lenA:lenB;//数组c长度为a，b较长长度。 
	int carry=0,t=0;
	c[lenC]='';			//字符串以''做结束标志。 
	for(int i=lenC-1;i>=0;i--){
		lenA--;
		lenB--;
		if(lenA>=0&&lenB>=0){
			t=a[lenA]-'0'+b[lenB]-'0'+carry;carry=0;//a,b数组的末位开始相加，取其数值做判断加完进位后carry置0 
			if(t>9){
			c[i]=t-10+'0';//数值大于10 ，取其个位，进位carry置1. 
			carry=1;
			}
			else c[i]=t+'0';			
			}
			
		if(lenA>=0&&lenB<0){// a数组位数较长，将a数组多余字符赋给c。 
			t=a[lenA]-'0'+carry;carry=0;
			if(t>9){
				c[i]=t-10+'0';
				carry=1;
			}
			else c[i]=t+'0';
			}
		if(lenB>=0&&lenA<0){//b数组较长 
			t=b[lenB]-'0'+carry;carry=0;
			if(t>9){
				c[i]=t-10+'0';
				carry=1;
			}
			else c[i]=t+'0';
		}
	}	
}

（编辑：李大同）

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