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  • 大数求余/取模

     1 /**2016中国大学生程序设计网络赛赛HDU
     2    1001大数求余
     3 */
     4 
     5 #include "iostream"
     6 #include "cstdio"
     7 #include "cstring"
     8 using namespace std;
     9  char s[100000002];
    10 int main() {
    11 
    12   int t=1;
    13   while(~scanf("%s", &s)) {
    14 
    15     int m = 0,n=0;
    16     for(int i = 0; i <strlen(s); i++)
    17      {
    18        m = (m*10+s[i]-'0')%73;
    19        n = (n*10+s[i]-'0')%137;
    20      }
    21 
    22      if(m==0&&m==0)
    23      {
    24         printf("Case #%d: YES
    ",t);
    25      }else{
    26         printf("Case #%d: NO
    ",t);
    27      }
    28     t++;
    29   }
    30   return 0;
    31 }

    ///大整数取模
    /**
    输入正整数n m输出n mod m n<=10^100 m<=10^9
    */

    将大整数写成自左向右形式:1234=((1*10+2)*10+3)*10+4

     1 #include "cstdio"
     2 #include "cstring"
     3 #include "cstdlib"
     4 using namespace std;
     5 int main()
     6 {
     7     char n[101];
     8     int m;
     9     while(~scanf("%s%d",n,&m)&&m)
    10     {
    11         int len=strlen(n);
    12         int ans=0;
    13         for(int i=0;i<len;i++)
    14             ans=(int)(((long long)ans*10+n[i]-'0')%m);
    15         printf("%d
    ",ans);
    16     }
    17     return 0;
    18 }
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  • 原文地址:https://www.cnblogs.com/kimsimple/p/5779035.html
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