hdu 1002 A + B Problem II (大数相加)

发布时间：2020-12-14 03:32:52 所属栏目：大数据来源：网络整理

导读：链接： hdu 1002 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 tes

链接：hdu 1002

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

解析：本题就是模拟的数学算数相加，可以从个位开始相加，超过10就先前进位

AC代码：

#include<stdio.h>
#include<string.h>
char s[1010],s1[1010],s2[1010];
void add()
{
    int i,j,k,m,n;
    m=strlen(s1);
    n=strlen(s2);
    memset(s,sizeof(s));
    if(m>n)
        k=m-1;
    else
        k=n-1;
    for(i=m-1,j=n-1;i>=0||j>=0;i--,j--){
        if(i>=0&&j>=0)
            s[k]+=s1[i]+s2[j]-48;
        else if(i>=0&&j<0)
            s[k]+=s1[i];
        else if(i<0&&j>=0)
            s[k]+=s2[j];
        if(s[k]>'9'&&k){ //k为0时，不能往数组前一位进位，要另外判断
            s[k]-=10;
            s[k-1]++;
        }
        k--;
    }
}
int main()
{
    int i,T;
    scanf("%d",&T);
    for(i=1;i<=T;i++){
        scanf("%s%s",s1,s2);
        printf("Case %d:n%s + %s = ",i,s2);
        add();
        if(s[0]>'9'){   //最高位为0时
            s[0]-=10;
            printf("1");
        }
        printf("%sn",s);
        if(i!=T)
            printf("n");
    }
    return 0;
}

（编辑：李大同）

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