HDOJ Integer Inquiry (大数累加)
发布时间:2020-12-14 02:27:36 所属栏目:大数据 来源:网络整理
导读:Integer Inquiry Time Limit : 2000/1000ms (Java/Other)???Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 24???Accepted Submission(s) : 7 Problem Description One of the first users of BIT's new supercomputer was Chip Diller. H
Integer InquiryTime Limit : 2000/1000ms (Java/Other)???Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 24???Accepted Submission(s) : 7
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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ac代码: #include<stdio.h> #include<string.h> char s[1005]; int num[1006]; int main() { int t,i,j; while(scanf("%d",&t)!=EOF) { while(t--) { memset(num,sizeof(num)); while(scanf("%s",s)&&s[0]!='0') { int c=1; int lens=strlen(s); for(j=lens-1;j>=0;j--) { num[c]+=s[j]-'0'; num[c+1]+=num[c]/10; num[c]%=10; c++; } } int bz=0; for(i=1005;i>1;i--) { if(num[i]==0&&bz==0) continue; else { bz=1; printf("%d",num[i]); } } printf("%dn",num[1]); 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! |