HDU-1002-大数相加

最新推荐文章于 2019-06-12 17:10:05 发布

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本文介绍了一个名为“A+B Problem II”的编程题目，该题目要求计算两个大整数的和，并提供了一种使用C语言实现的解决方案。输入包含多个测试案例，每个案例由两个大整数组成，需要编写程序计算它们的和并按特定格式输出结果。

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 317183 Accepted Submission(s): 61609

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 = 1111111111111111110

Author

Ignatius.L

#include<stdio.h>
#include<string.h>
#define MAX 1001
void largenumber(char a[],char b[],char d[])
{
    int i,j;
    int t=0;//补位
    int A[MAX]={0};
    int B[MAX]={0};
    char c[MAX];
int a_length,b_length;
int max_size;
a_length=strlen(a);
b_length=strlen(b);
max_size=(a_length>b_length)?a_length:b_length;//求最大的长度
for(i=0;i<max_size;i++)//倒置字符串用int保存，多余位置为0
{
(a_length>i)?A[i]=a[a_length-i-1]-'0':A[i]=0;
(b_length>i)?B[i]=b[b_length-i-1]-'0':B[i]=0;
}
for(i=0;i<max_size;i++)
{
if((A[i]+B[i]+t)>=10)//加了补位大于10
{
c[i]=(A[i]+B[i]+t)%10+'0';
t=1;
}
else
{
c[i]=(A[i]+B[i]+t)+'0';
t=0;
}

}
if(t==1){c[i]=t+'0';c[i+1]='\0';}//对于最后位补位的判断
else
{
c[i]='\0';
}
for(i=0;i<strlen(c);i++)//倒置
d[i]=c[strlen(c)-i-1];
d[i]='\0';

}
int main()
{
    char a[MAX],b[MAX],c[MAX];
    int n;
    int count=0;
    scanf("%d",&n);
    while(n--)
    {
        count++;
    scanf("%s",a);
    scanf("%s",b);
largenumber(a,b,c);
printf("Case %d:\n",count);
printf("%s + %s = %s",a,b,c);
if(n!=0) printf("\n\n");
else printf("\n");
    }

return 0;
}