HDU 1047--Integer Inquiry 【大数】
发布时间:2020-12-14 02:27:31 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): 15194????Accepted Submission(s): 3907 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): 15194????Accepted Submission(s): 3907
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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#include <cstdio> #include <cstring> int a[1100]; void add(char * str){ int b[1100]; memset(b,sizeof(b)); int len = strlen(str); int i; int k = 0; for(i = len - 1; i >= 0; i--) b[k++] = str[i] - '0'; for( i = 0; i < 1100 ; i++){ a[i] += b[i]; if(a[i] > 9){ a[i] = a[i] % 10; a[i + 1]++; } } } int main (){ int T; scanf("%d",&T); while(T--){ char s[1100]; memset(a,sizeof(a)); while(scanf("%s",s)){ if(!strcmp(s,"0")) break; add(s); } int i; for(i = 1100; i >= 0 && !a[i]; i--); if(i < 0) //处理直接输入0的情况,此时输出 0 printf("0n"); else { for(; i >= 0; i--) printf("%d",a[i]); printf("n"); } if(T) 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! |