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hdu 1002 a + b problem ii

admin 11月 13, 2020 0

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 257918 Accepted Submission(s): 49865

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

1
2
3

2
1 2
112233445566778899 998877665544332211

Sample Output

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

Idea

原题传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1002
新手必做题，简单的高精度加法，写了两个代码都AC了，第一个快些，第二个好理解些

Code1


#include <stdlib.h>
#include <string.h>
#define MAX 1005

int (void)
{
    int n, i = 1, n1, n2, k, j, p, q, x, m;
    char a[MAX], b[MAX], c[MAX];

    
    //freopen("output.txt","w", stdout);
    scanf("%d", &n);
    while(n)
    {
        scanf("%s %s", a, b);
        printf("Case %d:n%s + %s = ", i, a, b);
        n1 = strlen(a) - 1;
        n2 = strlen(b) - 1;
        k = x = 0;
        while( n1 >= 0 || n2 >= 0)
        {
            p = n1 >= 0 ? a[n1--] - '0' : 0;
            q = n2 >= 0 ? b[n2--] - '0' : 0;
            m = p + q + x;
            x = m / 10;
            m = m % 10;
            c[k++] = m + '0';
        }
        for(j = k - 1; j >=0; j--)
            printf("%c", c[j]);
        i++;
        if(n != 1) printf("nn");
        else printf("n");
        n--;
    }
    return 0;
}

Code2


#include <stdlib.h>
#include <string.h>
#define MAX 1015

void Add(char *a, char *b, char *c);
int (void)
{
    int n;
    int i;
    char s1[MAX], s2[MAX], s3[MAX];

    
    //freopen("output.txt","w", stdout);
    scanf("%d", &n);
    //while(getchar() != 'n') continue;
    for(i = 1; i <= n; i++) { scanf("%s %s", s1, s2); Add(s1, s2, s3); printf("Case %d:n%s + %s = %s", i, s1, s2, s3); if(i == n) printf("n"); else printf("nn"); } return 0; } void Add(char *a, char *b, char *c) { int i = strlen(a) - 1, j = strlen(b) - 1, k = 0; int x = 0, m = 0; char t[MAX]; int p, q; while( i >= 0 || j >= 0)
    {
        p = i >= 0 ? a[i--] - '0' : 0;
        q = j >= 0 ? b[j--] - '0' : 0;
        m = p + q + x;
        x = m / 10;
        m = m % 10;
        t[k++] = m + '0';
    }
    for(i = 0; i < k; i++)
        c[i] = t[k - i - 1];
    c[k] = '