zoukankan      html  css  js  c++  java
  • 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;
    }


  • 相关阅读:
    Orchard学习 02、orchard 路由
    Orchard学习 01、orchard日志
    golang限制协程的最大开启数
    go爬取博客园
    go xpath 添加头部解析
    goadmin文档
    使用GoAdmin极速搭建golang应用管理后台
    Python GUI之tkinter窗口视窗教程大集合(看这篇就够了)转
    自定义推荐央视
    python爬虫 xpath
  • 原文地址:https://www.cnblogs.com/Tovi/p/6194761.html
Copyright © 2011-2022 走看看