poj1503Integer Inquiry

最新推荐文章于 2019-03-23 17:24:53 发布

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#ACM #算法 #poj1053Integer Inqui #大整数相加

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本文介绍了一个关于VeryLongInteger求和的问题，通过读取一系列不超过100个字符的非负整数，并输出这些整数的总和。使用C语言实现，主要涉及字符串处理和大数运算。

原题链接

Integer Inquiry

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 33777		Accepted: 13157

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

题目和代码都不难就不多说直接看代码吧

#include
  
   
#include
   
    
#define MAXN 105
int b[MAXN];
char a[MAXN];
int main(){
	int len,i,j,k;
	while(scanf("%s",a)!=EOF){
		if(strcmp(a,"0")==0){
			break;
		}
		else{
			len=strlen(a);
			for(i=0;i
    
     =10){
					b[j+1]=b[j+1]+b[j]/10;
					b[j]=b[j]%10;
				}
			}	
		}
	}
	k=MAXN;
	while(b[--k]==0);
	while(k>=0){
		printf("%d",b[k--]);
	}
}