hdu 1212 Big Number(数论)
Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3256 Accepted Submission(s): 2219
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.
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
Recommend
Eddy
思路:一个数模上另一个数等于10进制数展开的模相加的模。
如 532 mod 7 =(500%7+30%7+2%7)%7;
当然还有a*b mod c=(a mod c+b mod c)mod c;
如35 mod 3=((5%3)*(7%3))%3
知道这些这题就很容易了。
#include<iostream> #include<cstring> using namespace std; const int mm=100020; char s[mm]; int ss; int main() { while(cin>>s>>ss) { int len=strlen(s); int ans=0; for(int i=0;i<len;i++) { ans=ans*10+s[i]-'0'; ans%=ss; } cout<<ans<<"\n"; } }
The article write by nealgavin