zoukankan      html  css  js  c++  java
  • Poj 1503 Integer Inquiry

    1.链接地址:

    http://poj.org/problem?id=1503

    2.题目:

    Integer Inquiry
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 28115   Accepted: 10925

    Description

    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

    3.思路:

    4.代码:

     1 #include "stdio.h"
     2 #include "string.h"
     3 #define NUM 102
     4 int a[NUM];
     5 int main()
     6 {
     7     char b[102];
     8     int len;
     9     int max=0;
    10     int i;
    11     gets(b);
    12     while(b[0]-'0'!=0 || strlen(b)!=1)
    13     {
    14         len = strlen(b);
    15         for(i=len-1;i>=0;i--)
    16         {
    17             a[len-1-i]+=b[i]-'0';
    18             if(a[len-1-i]>=10) {a[len-1-i+1]++;a[len-1-i]-=10;}
    19         }
    20         i=len;
    21         while(a[i]>=10){a[i+1]++;a[i]-=10;i++;}
    22         gets(b);        
    23     }
    24     i=NUM-1;
    25     while(a[i]==0)i--;
    26     while(i>=0)
    27     {
    28         printf("%c",a[i]+'0');
    29         i--;
    30     }
    31    return 0;
    32 }
  • 相关阅读:
    Java编程基础
    Python开发【第十四篇】:Python操作MySQL
    MySQL(二)
    MySQL(一)
    Python之路【第五篇】:面向对象及相关
    Python开发【第四篇】:Python基础之函数
    Python开发【第三篇】:Python基本数据类型
    等保测评备案流程?备案资料有哪些?
    xls/csv文件转换成dbf文件
    csv 转换为DBF文件的方法
  • 原文地址:https://www.cnblogs.com/mobileliker/p/3556808.html
Copyright © 2011-2022 走看看