• HDU1002 A + B Problem II【大数】


    A + B Problem II

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 371486    Accepted Submission(s): 72398


    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

    问题链接HDU1002 A + B Problem II

    问题简述:参见上文。

    问题分析这个问题是要计算两个数相加的结果,问题是每个数的长度不超过1000,也就是数可能非常大,所有需要用大数加法来做。

    程序说明:程序中处处都是基本的套路,需要熟练掌握。

    题记:程序需要写得简洁易懂。


    参考链接:(略)


    AC的C++语言程序如下:

    /* HDU1002 A + B Problem II */
    
    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    
    using namespace std;
    
    const int BASE = 10;
    const int N = 1000;
    char sa[N+1], sb[N+1];
    int a[N+1], b[N+1];
    
    int main()
    {
        int t;
    
        scanf("%d", &t);
    
        for(int k=1; k<=t; k++) {
            scanf("%s%s", sa, sb);
    
            // 字符串转数值数组
            memset(a, 0, sizeof(a));
            memset(b, 0, sizeof(b));
            int alen = strlen(sa);
            for(int i=alen-1,j=0; i>=0; i--,j++)
                a[j] = sa[i] - '0';
            int blen = strlen(sb);
            for(int i=blen-1,j=0; i>=0; i--,j++)
                b[j] = sb[i] - '0';
    
            // 相加(需要考虑进位问题)
            int len = max(alen, blen);
            int carry = 0;
            for(int i=0; i<len; i++) {
                a[i] += b[i] + carry;
                carry = a[i] / BASE;
                a[i] %= BASE;
            }
            if(carry > 0)
                a[len++] = carry;
    
            // 输出结果(需要注意输出格式:隔行输出)
            if(k != 1)
                printf("
    ");
            printf("Case %d:
    ", k);
            printf ("%s + %s = ", sa, sb);
            for(int i=len-1; i>=0; i--)
                printf("%d", a[i]);
            printf("
    ");
        }
    
        return 0;
    }





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  • 原文地址:https://www.cnblogs.com/tigerisland/p/7563540.html
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