HDU 1212 Big Number（大数取模）
发布时间：2021-03-11 17:22:48 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): 7063????Accepted Submission(s): 4866 Problem Description As we know,Big Number is always troublesome. But it's really imp
|
Big NumberTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7063????Accepted Submission(s): 4866 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 2 3 12 7 152455856554521 3250? Sample Output 2 5 1521? Author Ignatius.L ? Source 杭电ACM省赛集训队选拔赛之热身赛 ? Recommend Eddy???|???We have carefully selected several similar problems for you:?? 1215? 1211? 1210? 1214? 1002? 代码： #include<stdio.h>
#include<string.h>
using namespace std;
int mod(char n1[],int n)
{
int len=strlen(n1);
int temp=0;
for(int i=0;i<len;i++)
{
temp=temp*10+n1[i]-'0';
temp=temp%n;
}
return temp;
}
int main()
{
char s[1000000];
int num;
while(~scanf("%s%d*c",&s,&num))
{
printf("%dn",mod(s,num));
}
}
（编辑：我爱故事小小网_铜陵站长网） 【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! |
浙公网安备 33038102330570号