博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
Integer Inquiry(UVA—424)
阅读量:3958 次
发布时间:2019-05-24

本文共 1564 字,大约阅读时间需要 5 分钟。

题目:

One of the first users of BIT’s new supercomputer was Chip Diller. He extended his exploration ofpowers 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 Skyapartments 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 (noVeryLongInteger 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

1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678900

<font color=red Sample Output

370370367037037036703703703670

解题思路:这个题是高精度的问题,多个大数相加的结果.

程序代码:

#include
#include
int a[50000],b[50000],c[55000];char s[200];int main(){
int n,m,i,j,k; memset(a,0,sizeof(a)); scanf("%s",s); m=strlen(s); j=0; for(i=m-1;i>=0;i--) a[j++]=s[i]-'0'; while(scanf("%s",s)!=EOF) {
memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); if(s[0]=='0') break; n=strlen(s); j=0; for(i=n-1;i>=0;i--) b[j++]=s[i]-'0'; if(n>m) k=n; else k=m; for(i=0;i
=10) {
c[i+1]++; c[i]=c[i]%10; } } if(c[k]!=0) k++; for(i=0;i
=0;i--) printf("%d",a[i]); printf("\n"); return 0; }

转载地址:http://fexzi.baihongyu.com/

你可能感兴趣的文章
(八) 正则表达式
查看>>
一.JavaScript 基础
查看>>
7.ECMAScript 继承
查看>>
HTML DOM
查看>>
AJAX 基础
查看>>
JSON 基础
查看>>
J2EE监听器Listener接口大全[转]
查看>>
cookie、session、sessionid 与jsessionid[转]
查看>>
常见Oracle HINT的用法
查看>>
JAVA中各类CACHE机制实现的比较 [转]
查看>>
PL/SQL Developer技巧
查看>>
3-python之PyCharm如何新建项目
查看>>
15-python之while循环嵌套应用场景
查看>>
17-python之for循环
查看>>
18-python之while循环,for循环与else的配合
查看>>
19-python之字符串简单介绍
查看>>
20-python之切片详细介绍
查看>>
P24-c++类继承-01详细的例子演示继承的好处
查看>>
P8-c++对象和类-01默认构造函数详解
查看>>
P1-c++函数详解-01函数的默认参数
查看>>