HDU1047 Integer Inquiry

最新推荐文章于 2019-09-02 19:24:20 发布

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本文介绍了一个关于超长整数求和的问题，输入由最多100行组成，每行包含一个非常长的整数，长度不超过100个字符，任务是计算这些整数的总和。文章提供了C++代码实现，展示了如何处理和计算这些超长整数。

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

11234567890123456789012345678901234567890123456789012345678901234567890123456789012345678900

Sample Output

370370367037037036703703703670

//1002题的改版，注意下格式就好了

#include <iostream>#include <cstdio>#include <string.h>using namespace std;#define N 105int main(){    void add(char a[],char b[],char c[]);    char a[N],b[N],sum[N];    int n;    cin >> n;    getchar();    while(n--)    {        strcpy(sum,"0");        while(1)        {            scanf("%s",a);            if(!strcmp(a,"0"))            break;            add(a,sum,sum);        }        printf("%s\n",sum);        if(n)        cout << endl;    }    return 0;}void add(char a[],char b[],char c[]){    int i,e,d;    int n,m;    char temp;    n=strlen(a);    m=strlen(b);    for(i=0; i<n/2; i++)    {        temp=a[i];        a[i]=a[n-1-i];        a[n-1-i]=temp;    }//将a倒置，另外倒置可以写成另外的函数以便节省代码量，后面多次调用    for(i=0; i<m/2; i++)    {        temp=b[i];        b[i]=b[m-1-i];        b[m-1-i]=temp;    }//将b倒置    e=0;    for(i=0; i<n&&i<m; i++)    {        d=a[i]-'0'+b[i]-'0'+e;        e=d/10;        c[i]=d%10+'0';    }//进行模拟加法    if(i<m)    {        for(; i<m; i++)        {            d=b[i]-'0'+e;            e=d/10;            c[i]=d%10+'0';        }    }    else    {        for(; i<n; i++)        {            d=a[i]-'0'+e;            e=d/10;            c[i]=d%10+'0';        }    }//如果有一个数的位数较多，特殊处理    if(e) c[i++]=e+'0';//处理最后一位可以进位的情况    c[i]=0;    n=i;    for(i=0; i<n/2; i++)    {        temp=c[i];        c[i]=c[n-1-i];        c[n-1-i]=temp;    }//倒置}