424 Integer Inquiry (Integer query)

Integer Inquiry

One of the first users of BIT's new supercomputer was Chip Diller. He extended his activation of powers of 3 to go from 0 to 333 and he got Ed 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 in a single zero on a line by itself.

Output
Your program shocould output the sum of the VeryLongIntegers given in the input.

Sample Input

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0
Sample Output

370370367037037036703703703670

 

 

 

Question:

The sum of big integers is actually like the vertical sum of a primary school. The sum of one digit and one digit is enough.

Note: After WA twice, you will find that a line break is required after the final output !!!

Later, it was found that it was not necessary to judge that the last row was 0, and the EOF could be read all the time.

 

The following is a code for two different ideas:

Code 1:

Read all rows and then calculate the large number.

 

#include <stdio.h>#include <stdlib.h>#include <string.h>#define INF 1e9char str[105][105];int num[105][105];int ans[10010];int main(){    int i, j;    int len, count, max = -INF, flag;    i = 0;    while(~scanf("%s", str[i])){        if(str[i][0] == '0')  break;        i++;    }    count = i;    memset(num, 0, sizeof(num));    for(i = 0; i<count; i++){        len = strlen(str[i]);        for(j = 0; j<len; j++){            num[i][len-j] = (str[i][j]-'0');        }        if(max<len)            max = len;    }    memset(ans, 0, sizeof(ans));    for(i = 0; i<count; i++){        for(j = 1; j<=max; j++){            ans[j] += num[i][j];        }    }    for(i = 1; i<10010;i++){        ans[i] += ans[i-1]/10;        ans[i-1] %= 10;    }    flag = 0;    for(i=10009; i>0; i--){        if(flag){            printf("%d", ans[i]);        }        else if(ans[i]){            flag = 1;            printf("%d", ans[i]);        }    }    printf("\n");    return 0;}

 

Code 2:
If a single string is not read, a large number is added at a time.

 

#include <stdio.h>#include <stdlib.h>#include <math.h>#include <string.h>char str[100];int num[100];int ans[10000];int main(){    int i, j;    int len, flag;    memset(ans, 0, sizeof(ans));    while(~scanf("%s", str)){        len = strlen(str);        memset(num, 0, sizeof(num));        for(i = 0; i<len; i++)            num[len-i] = str[i]-'0';        for(i = 1; i<=len; i++)            ans[i] += num[i];        for(i = 1; i<10000; i++){            ans[i] += (ans[i-1]/10);            ans[i-1] %= 10;        }    }    flag = 0;    for(i=10000-1; i>0; i--){        if(flag)            printf("%d", ans[i]);        else if(ans[i]){            printf("%d", ans[i]);            flag = 1;        }    }    printf("\n");    return 0;}