HDU - 1047 - Integer Inquiry (大数高精度)
发布时间:2020-12-14 02:43:40 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): 13969????Accepted Submission(s): 3523 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): 13969????Accepted Submission(s): 3523
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
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简单大数题。。 AC代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; char a[110]; int ans[110]; int tmp[110]; void fun(char a[],int tmp[]) { int len = strlen(a); for(int i = len - 1; i >= 0 ; i--) { tmp[len - 1 - i] = a[i] - '0'; } } void add(int ans[],int tmp[]) { int up = 0; for(int i = 0; i < 110; i++) { int t = ans[i] + tmp[i] + up; ans[i] = t % 10; up = t / 10; } } void print(int ans[]) { int flag = 0; for(int i = 109; i >= 0; i--) { if(ans[i] > 0) { flag = 1; printf("%d",ans[i]); } else if(flag && ans[i] == 0) printf("0"); } if(flag == 0) printf("0"); printf("n"); } int main() { int n; scanf("%d",&n); while(n--) { scanf("%s",a); if(strcmp(a,"0") == 0) { //没判断只有一个0的情况,无语的WA了2次╮(╯▽╰)╭ printf("0n"); if(n) printf("n"); continue; } memset(ans,sizeof(ans)); fun(a,ans); while(1) { scanf("%s",a); if(strcmp(a,"0") == 0) break; memset(tmp,sizeof(tmp)); fun(a,tmp); add(ans,tmp); } print(ans); if(n) 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! |