poj 1503 Integer Inquiry（多個大數相加）

標籤：poj   大數   多個大數相加   

題目連結：http://poj.org/problem?id=1503

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

1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678900

Sample Output

370370367037037036703703703670

Source

East Central North America 1996

題意：就是給出多個大數，求它們的和！

下面給出兩種代碼：

代碼一：

#include <cstdio>#include <cstring>const int MAXN = 117;int main(){    char s[MAXN];    int sum[MAXN] = {0};    int i, j;    while(gets(s))    {        int len = strlen(s);        if(s[0] == '0' && len == 1)            break;        for(i = 110, j = len-1; j >= 0; i--, j--)        {            sum[i] += s[j]-'0';        }    }    for(i = 110; i > 0; i--)    {        sum[i-1] += sum[i] / 10;        sum[i] %= 10;    }    for(i = 0; sum[i] == 0 && i < 111; i++)    {        if(i == 111)//意味著全為零        {            printf("0\n");        }    }    for( ; i < 111; i++)    {        printf("%d",sum[i]);    }    printf("\n");    return 0;}

代碼二：

#include <cstdio>#include <cstring>const int MAXN = 117;int main(){    char s[MAXN][MAXN];    int maxx = -1, r = 0;    for(int i = 0; ; i++)    {        gets(s[i]);        int len = strlen(s[i]);        if(len > maxx)//尋找最長的長度            maxx = len;        if(s[i][0] == '0' && len == 1)            break;        r++;    }    int c[MAXN], l = 0;    int p = 0;//進位    for(int i = maxx-1; i >= 0; i--)    {        int sum = 0;        for(int j = 0; j < r; j++)        {            sum+=s[j][i]-'0';        }        sum += p;        if(sum > 9)        {            p = sum/10;            sum %=10;        }        else            p = 0;        c[l++] = sum;        if(i == 0 && p != 0)            c[l++] = p;    }    for(int i = l-1; i >= 0; i--)    {        printf("%d",c[i]);    }    printf("\n");    return 0;}