HDOJ 1212 Big Number(大数版同余定理)
发布时间:2020-12-14 02:27:34 所属栏目:大数据 来源:网络整理
导读:Big Number Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5916????Accepted Submission(s): 4135 Problem Description As we know,Big Number is always troublesome. But it's really impor
Big NumberTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Problem Description
As we know,Big Number is always troublesome. But it's really important in our ACM. And today,your task is to write a program to calculate A mod B.
To make the problem easier,I promise that B will be smaller than 100000. Is it too hard? No,I work it out in 10 minutes,and my program contains less than 25 lines.
?
Input
The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000,and B will be smaller than 100000. Process to the end of file.
?
Output
For each test case,you have to ouput the result of A mod B.
?
Sample Input
?
Sample Output
?
乍一看还以为是大数相减,通过不断相减得到小于n的值,即余数。准备敲时,感觉会超时,放弃了。 ? 正解则是使用同余定理不断对s取模。 ? 这一题用了以下两个同余定理公式: ????????????? (A?+?B)?mod?M?=?(?A?mod?M?+?B?mod?M?)?mod?M ??????????????(A?*?B)?mod?M?=?((A?mod?M)?*(?B?mod?M))?mod?M??
? #include<cstdio> #include<cstring> int main() { int n,len,i,ans; char str[1010]; while(scanf("%s%d",str,&n)!=EOF) { len=strlen(str); ans=0; for(i=0;i<len;i++) { ans=ans*10+(str[i]-'0'); ans=ans%n; } printf("%dn",ans); } 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! |
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