HDOJ 5832 A water problem(高精度 大数取模)
发布时间:2020-12-14 01:36:31 所属栏目:大数据 来源:网络整理
导读:A water problem Time Limit: 5000/2500 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 826????Accepted Submission(s): 428 Problem Description Two planets named Haha and Xixi in the universe and they were c
A water problemTime Limit: 5000/2500 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 826????Accepted Submission(s): 428
Problem Description
Two planets named Haha and Xixi in the universe and they were created with the universe beginning.
There is Now you know the days
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Input
There are several test cases(about
For each test,we have a line with an only integer
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Output
For the i-th test case,output Case #i:,then output "YES" or "NO" for the answer.
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Sample Input
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Sample Output
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Author
UESTC
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Source
2016中国大学生程序设计竞赛 - 网络选拔赛?
思路:
这道题的题意很清楚,就是让你求73和137的的公倍数。因为73和137的最小公倍数是10001,所以其实也就是求这个数能不能模10001余0。但是由于输入的数是10000000位的数,用long long来装是肯定不行的,所以考虑用高精度。套用一个大数取余的模板
就可以。这里选择手动大数取模。
代码:
#include<iostream> #include<string> #include<cstdio> int const maxn=10000000+5; char str[maxn]; using namespace std; int main(){ //freopen("in.txt","r",stdin); int t=0; while(scanf("%s",str)!=EOF) { t++; int i; int rem=0; for(i=0;str[i];i++){ rem=rem*10+str[i]-'0'; rem=rem%10001; } printf("Case #%d: ",t); if(rem==0) printf("YESn"); else printf("NOn"); } } (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! |