zoukankan      html  css  js  c++  java
  • UVA_Integer Inquiry

    A - Integer Inquiry
    Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu
    Submit Status

    Description

    Download as PDF
     

     

     Integer Inquiry 

    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 VeryLongInteger. 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

    题意:给你n行大数;求它们的和;
    直接套用刘汝佳的模板;


    代码:
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstdlib>
     4 #include <cstring>
     5 #define UNIT 10
     6 
     7 using namespace std;
     8 
     9 struct Bignum
    10 {
    11     int val[105];
    12     int len;
    13 
    14     Bignum ()
    15     {
    16         memset (val, 0, sizeof(val));
    17         len = 1;
    18     }
    19 
    20     Bignum operator + (const Bignum &a) const//大数加大数
    21     {
    22         Bignum x = a;
    23         int L;
    24         L = a.len > len ? a.len : len;
    25         for (int i = 0; i < L; ++i)
    26         {
    27             x.val[i] += val[i];
    28             if (x.val[i] >= UNIT)
    29             {
    30                 x.val[i+1]++;
    31                 x.val[i] -= UNIT;
    32             }
    33         }
    34         if (x.val[L] != 0)
    35             x.len = L+1;
    36         else
    37             x.len = L;
    38         return x;
    39     }
    40 
    41 };
    42 int main()
    43 {
    44    // freopen("ACM.txt","r",stdin);
    45     char a[105];
    46     Bignum x,ans;
    47     for(;;)
    48     {
    49         Bignum(x);
    50         scanf("%s",a);
    51         int L=strlen(a);
    52         x.len=0;
    53         for(int i=L-1;i>=0;i--)
    54         {
    55             x.val[x.len++]=a[i]-'0';
    56         }
    57         if(x.len==1&&x.val[0]==0) break;
    58         ans=ans+x;
    59 
    60     }
    61     for(int i=ans.len-1;i>=0;i--)
    62     cout<<ans.val[i];
    63     cout<<endl;
    64     return 0;
    65 }
    View Code





     
  • 相关阅读:
    pwnable.kr login之write up
    安装ubuntu16.4后
    HTML资源定位器-URL
    【实验吧】编程循环&&求底运算
    【SQL注入】mysql中information_schema详解
    C#之BackgroundWorker从简单入门到深入精通的用法总结
    C#使用NPOI对Excel文档进行读、写、导入、导出等操作的dll最新版2.5.1+2.3.0
    Visual Studio中Debug与Release以及x86、x64、Any CPU的区别
    C# 使用BackgroundWorker例子及注意点
    C# BackgroundWorker组件学习入门介绍
  • 原文地址:https://www.cnblogs.com/M-D-LUFFI/p/4050143.html
Copyright © 2011-2022 走看看