HDU-5832 HDU-1212/2016网络选拔赛/大数取余

本文介绍了解决ACM竞赛中涉及大数运算问题的方法,包括如何计算两个行星是否处于其年周期的第一天及大数取模问题的解决思路。通过具体的样例输入输出展示了算法的有效性。

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A water problem

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 326    Accepted Submission(s): 180


Problem Description
Two planets named Haha and Xixi in the universe and they were created with the universe beginning.

There is 73 days in Xixi a year and 137 days in Haha a year.

Now you know the days N after Big Bang, you need to answer whether it is the first day in a year about the two planets.
 

Input
There are several test cases(about 5 huge test cases).

For each test, we have a line with an only integer N(0N) , the length of N is up to 10000000 .
 

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
 

Source

Cmod(A*B)==0  =CmodA==0&&CmodB==0
 
   C
---------- =integer
  A*B
#include <iostream>
using namespace std;
#define N 12000000
#define INF 1<<30
char s[N];
int main()
{
    int k=0,len,x,d,i;
    while(scanf("%s", s) != EOF)
    {
        k++;
        len = strlen(s);
        x=10001;
        d=0;
        for(i=0;i<len;i++)
            d=(d*10+(s[i]-'0')%x)%x;

        if(d==0)
            printf("Case #%d: YES\n",k);
        else
            printf("Case #%d: NO\n",k);
    }
    return 0;
}

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7409    Accepted Submission(s): 5124


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

#include <iostream>
#include<string.h>
using namespace std ;
#define N 1005
char s [N ];
int main()
{
  int k = 0 ,len ,x ,d ,i ;
   while(scanf ( "%s" ,s )!= EOF ) {
        cin >>x ;
        k ++;
        len = strlen (s );
        d = 0 ;
      for(i = 0 ;i <len ;i ++)
            d =(d * 10 +(s [i ]- '0' )%x )%x ;
        printf ( "%d\n" ,d );
   }
  return 0 ;
}
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