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hdu 1211

2024-07-02 03:34:18   225人阅读

这题就是 要你找出一个ASCII 的值x使得  ：  x^e%n==num(当前输入的数，e条件已给出)

zsd:

1: ASCII0-255可以枚举

2:   =a^11   11=1011

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int pow2( int a, int b )
{
    int r = 1, base = a;
    while( b != 0 )
    {
        if( b % 2 )
            r *= base;
        base *= base;  //a^x x=1 2 4 8 16  也就是2^x
        b /= 2;
    }
    return r;
}

 3:(x*y)%d=(x%d)*(y%d)

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#include<iostream>
using namespace std;
bool funtion(int x,int e,int n,int num)
{// m^n % k
   int b = 1;
    while (e > 0)
    {
          if (e & 1)
             b = (b*x)%n;
          e = e >> 1 ;
          x = (x*x)%n;
    }
    if(b==num) return true;
    return false;
}
int main()
{
    int p,q,e,l,c,n;
    while(scanf("%d%d%d%d",&p,&q,&e,&l)!=EOF)
    {
        n=p*q;
        while(l--)
        {
            scanf("%d",&c);
            for(int i=0;i<=255;i++)
                if(funtion(i,e,n,c))
                {
                    printf("%c",i);
                    break;
                }
        }
        printf("\n");
    }
    return 0;
}

 4:貌似没有用到模线性方程

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#include<iostream>
      
using namespace std;
      
bool funtion(int x,int e,int n,int num)
      
{// m^n % k
      
   int b = 1;
      
    while (e > 0)
      
    {
      
          if (e & 1)
      
             b = (b*x)%n;
      
          e = e >> 1 ;
      
          x = (x*x)%n;
      
    }
      
    if(b==num) return true;
      
    return false;
      
}
      
int main()
      
{
      
    int p,q,e,l,c,n;
      
    while(scanf("%d%d%d%d",&p,&q,&e,&l)!=EOF)
      
    {
      
        n=p*q;
      
        while(l--)
      
        {
      
            scanf("%d",&c);
      
            for(int i=0;i<=255;i++)
      
                if(funtion(i,e,n,c))
      
                {
      
                    printf("%c",i);
      
                    break;
      
                }
      
        }
      
        printf("\n");
      
    }
      
    return 0;
      
}
 
 






























        

            
                
                  
                
                

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