zoukankan      html  css  js  c++  java
  • Integer Inquiry【大数的加法举例】

    Integer Inquiry
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 27730   Accepted: 10764

    Description

    题目链接:http://poj.org/problem?id=1503

    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

    Source

    题目描述:输入多个大数,求其和,输入以0结束。
    代码:
     1 #include<iostream>
     2 #include<string.h>
     3 #include<stdlib.h>
     4 using namespace std;
     5 int main()
     6 {
     7     char f1[110];
     8     int sum[110]={0},i,j;
     9     cin>>f1;
    10     while(strcmp(f1,"0")!=0)
    11     {
    12         int t=strlen(f1),f[110]={0};
    13         for(i=t-1,j=0;i>=0;i--,j++)
    14             f[j]=f1[i]-'0';
    15             int up=0;
    16             for(j=0;j<=105;j++)
    17             {
    18                 int s=f[j]+sum[j]+up;
    19                 sum[j]=s%10;
    20                 up=s/10;
    21             }
    22             cin>>f1;
    23     }
    24     for(i=105;i>=0;i--)
    25         if(sum[i]!=0)
    26         {
    27             while(i>=0)
    28             {
    29                 cout<<sum[i];
    30                 i--;
    31             }
    32             break;
    33         }
    34         cout<<endl;
    35         return 0;
    36 }
    37 
    38 
    39            
    View Code
  • 相关阅读:
    维克里拍卖 Vickrey auction
    弱占优策略--Weakly Dominant Strategy
    乱码电路(Garbled circuits)
    P和NP问题
    揭秘Facebook首个数据中心:全球15亿用户的账户信息都在这里
    数学符号“s.t.”的意义
    PKI系统深入介绍
    [转]公钥,私钥和数字签名这样最好理解
    Exif
    任我行 CRM 9.4
  • 原文地址:https://www.cnblogs.com/kuangdaoyizhimei/p/3392977.html
Copyright © 2011-2022 走看看