HDU 1047。多个大数相加
发布时间:2020-12-14 02:51:13 所属栏目:大数据 来源:网络整理
导读: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 D
Integer InquiryTime 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.)
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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.
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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.
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Sample Input
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Sample Output
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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 205 int 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,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("0n"); } 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,sizeof(a1));//一个案例结束后初始化a1数组。 } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |