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大数加法

admin 11月 13, 2020 0

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

mycode


#include<string.h>

int () {
    int i,j,k=0;
    char a[1001],b[1001];
    int c[1001]={0};
    scanf("%s%s",&a,&b);

    int lena=strlen(a);
    int lenb=strlen(b);
    int l=lena>lenb?lena+1:lenb+1;

    while(lena>0 && lenb>0) {
        if(c[k]+a[lena-1]-'0'+b[lenb-1]-'0'>9) {
            c[k]=c[k]+a[lena-1]-'0'+b[lenb-1]-'0'-10;
            c[k+1]++;
            k++;
            lena--;
            lenb--;
        } else {
            c[k]=a[lena-1]-'0'+b[lenb-1]-'0';
            k++;
            lena--;
            lenb--;
        }
    }

    if(lena>0) {
        c[k]=c[k]+a[lena-1]-'0';
        lena--;
        k++;
        while(lena>0){
            c[k++]=a[lena--]-'0';
        }
    } else if(lenb>0) {
        c[k]=c[k]+b[lenb-1]-'0';
        lenb--;
        k++;
        while(lenb>0){
            c[k++]=b[lenb--]-'0';
        }
    }
    if(c[k]!=0) {
        k++;
    }

    printf(" %s + %s = ",a,b);

    for(i=k-1; i>=0; i--)
        printf("%d",c[i]);
    printf("n");

    return 0;
}