HDU 5832：A water problem （大数整除）
发布时间：2021-01-24 03:32:19 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
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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. ? Sample Input 10001 0 333? Sample Output Case #1: YES Case #2: YES Case #3: NO? Author UESTC ? 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;
}
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