HDU 5832:A water problem (大数整除)
发布时间:2020-12-14 01:37:16 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): 412????Accepted Submission(s): 218 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): 412????Accepted Submission(s): 218
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?
?
Input
There are several test cases(about?
For each test,we have a line with an only integer?
?
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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题意:判断输入的数是否可以同时整除73与137,也就是能不能整除73与137的最小公倍数。
因为输入的数的长度是10000000以内,我们只能采用字符串来存储它,从ans初始化为0,最低位开始ans=(ans+s[i]-'0')%b;?计算到最后,如果ans?等于0,可以整除,否则,不可以整除。
AC代码:
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; char s[11000000]; int ans,b=73*137; int main() { for(int i=1; gets(s); i++) { ans=0; int l=strlen(s); for(int j=0; j<l; j++) ans=(ans*10+(s[j]-'0'))%b; if(!ans)printf("Case #%d: YESn",i); else printf("Case #%d: NOn",i); } 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! |