hdu1047 Integer Inquiry 多次大数相加
发布时间:2020-12-14 03:32:07 所属栏目:大数据 来源:网络整理
导读:转载请注明出处:http://blog.csdn.net/u012860063 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1047 Problem Description One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to g
转载请注明出处:http://blog.csdn.net/u012860063 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1047
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?
纯大树相加: 代码如下: #include <cstdio> #include <cstring> int sum[147]; char s[147]; void Add( char ss[]) { int len = strlen(ss); int z = 1; for(int i = len - 1 ; i >= 0 ; i-- ) { sum[z] +=(ss[i]-'0'); sum[z+1] +=sum[z]/10; sum[z]%=10; z++; } } int main() { int t; while(~scanf("%d",&t)) { while(t--) { memset(sum,sizeof(sum)); while(scanf("%s",s)&&s[0]!='0') { Add(s); } int flag = 0; for(int i = 147; i > 1 ; i-- ) { if(flag == 0 && sum[i] == 0) continue; else { flag = 1; printf("%d",sum[i]); } } printf("%dn",sum[1]); if(t != 0) printf("n"); } } 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! |