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  • UVA424高精度加法

    One of the first users of BIT’s new supercomputer was Chip Diller. He extended his exploration of
    powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.
    “This supercomputer is great,” remarked Chip. “I only wish Timothy were here to see these results.”
    (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky
    apartments on Third Street.)
    Input
    The input will consist of at most 100 lines of text, each of which contains a single VeryLongInte-
    ger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits (no
    VeryLongInteger will be negative).
    The final input line will contain a single zero on a line by itself.
    Output
    Your program should output the sum of the VeryLongIntegers given in the input.
    Sample Input
    123456789012345678901234567890
    123456789012345678901234567890
    123456789012345678901234567890
    0
    Sample Output
    370370367037037036703703703670
    #include <bits/stdc++.h>
    using namespace std;
    
    string add(string a, string b) {
        string c;
        int p1 = a.size(), p2 = b.size();
        int p = max(p1, p2), l = 0;
        while (1) {
            int k = (int)a[p1-1]-48+(int)b[p2-1]-48+l;
            char m = (char)48+k%10;
            c += m;
            l = k/10;
            p1 --, p2 --;
            if (p1 == 0 || p2 == 0) break;
        }
        if (p1 == 0 && p2 == 0) {
            if (l!=0) {
                char m = (char)48+l;
                c += m;
            }
        }
        else if (p1 != 0 && p2 == 0) {
            for (int i = p1; i>0; i--) {
                int k = (int)a[i-1]-48+l;
                char m = (char)48+k%10;
                c += m;
                l = k/10;
            }
            if (l!=0) {
                char m = (char)48+l;
                c += m;
            }
        }
        else if (p1 == 0 && p2 != 0) {
            for (int i = p2; i>0; i--) {
                int k = (int)b[i-1]-48+l;
                char m = (char)48+k%10;
                c += m;
                l = k/10;
            }
            if (l!=0) {
                char m = (char)48+l;
                c += m;
            }
        }
        string q;
        for (int i = 0; i<c.size(); i++)    q += c[c.size()-i-1];
        return q;
    }
    
    
    int main() {
        string s, a;
        a = "", s="0";
        int flag = 0;
        //cout <<add("9345", "900") <<endl;
        //cout << s << endl;
        while (cin >> a) {
            if (a[0] == '0')    break;
            
            s = add(s, a);
           // cout << s <<endl;
        }
        cout <<s <<endl;
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/Tovi/p/6194761.html
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