博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
hdu1047 Integer Inquiry 多次大数相加
阅读量:7106 次
发布时间:2019-06-28

本文共 1967 字,大约阅读时间需要 6 分钟。

转载请注明出处:

题目链接:

Problem 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.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
 
Sample Input
 
1 123456789012345678901234567890 123456789012345678901234567890 123456789012345678901234567890 0
 
Sample Output
 
370370367037037036703703703670
 
Source

纯大树相加:

代码例如以下:

#include 
#include
int sum[147]; char s[147];void Add( char ss[]){ int len = strlen(ss); int z = 1; for(int i = len - 1 ; i >= 0 ; i-- ) { sum[z] +=(ss[i]-'0'); sum[z+1] +=sum[z]/10; sum[z]%=10; z++; }}int main(){ int t; while(~scanf("%d",&t)) { while(t--) { memset(sum,0,sizeof(sum)); while(scanf("%s",s)&&s[0]!='0') { Add(s); } int flag = 0; for(int i = 147; i > 1 ; i-- ) { if(flag == 0 && sum[i] == 0) continue; else { flag = 1; printf("%d",sum[i]); } } printf("%d\n",sum[1]); if(t != 0) printf("\n"); } } return 0;}

你可能感兴趣的文章
第七周作业
查看>>
JQuery实现简单实用的气泡提示插件
查看>>
IDEA注册
查看>>
第四次实验
查看>>
12. MySQL简单使用
查看>>
GIS基础知识
查看>>
【java并发编程艺术学习】(五)第二章 java并发机制的底层实现原理 学习记录(三) 原子操作的实现原理学习...
查看>>
使用npm安装一些包失败了的看过来(npm国内镜像介绍)
查看>>
Goland中Redis的set求并集的错误处理
查看>>
Timer
查看>>
ComboBox
查看>>
C++ sort()函数和C qsort()函数用法总结
查看>>
【图像处理】工业相机原理详述 (转载)
查看>>
【分布式】Zookeeper应用场景
查看>>
【堆】
查看>>
Asp.net基础概念整理(一) Web应用程序和网站的区别
查看>>
[02-02 ]Java数据库链接范列
查看>>
一些常用的Bootstrap模板资源站
查看>>
taro 填坑之路(二)taro 通过事件监听 实现组件间传值
查看>>
数组操作
查看>>