HDU 1047。多个大数相加

Integer Inquiry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13496    Accepted Submission(s): 3390

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

 

Source

East Central North America 1996

 

题意，就是多个大数相加，以0为结束输入的标志，然后输出。坑点：（直接输入0时要输出0（因为这个WA了一次。），最后一个案例不需要输出两个空行。）

只要把两个大数的相加的会了，这个其实只是照搬。

贴个代码

 

#include <stdio.h>#include <string.h>#define N 205int main(){int n,lena,l,i,j,flag;char a[N];//字符数组int a1[N]={0};//  这个数组用来存最终的值    int a2[N];  //   这个数组用来存输入的值，下面不断的更新它，然后加到a1[N]这个数组上。    scanf("%d",&n);while(n--){flag=0;//用flag控制格式while(scanf("%s",a)!=EOF){flag++;if(a[0]=='0'){break;}memset(a2,0,sizeof(a2));  //每次都初始化它lena=strlen(a);for(i=lena-1,l=0;i>=0;i--)//直接存入a2[l++]=a[i]-'0';for(i=0;i<N;i++){a1[i]+=a2[i];  //加到a1的数组上去。if(a1[i]>=10)  //进位{a1[i]-=10;a1[i+1]++;}}}if(flag==1)  //如果flag为一说明输入就一个0，此时输出0。{printf("0\n");}else{for(i=N-1;i>=0;i--)if(a1[i])break;for(j=i;j>=0;j--)printf("%d",a1[j]);//输出a1数组。printf("\n");}if(n!=0)printf("\n");//格式控制。memset(a1,0,sizeof(a1));//一个案例结束后初始化a1数组。}return 0;}