hdu1370-Biorhythms

中國剩余定理            已知(n+d)%23=a;   (n+d)%28=b;   (n+c)%33=i         使33×28×a被23除余1，用33×28×6=5544；         使23×33×b被28除余1，用23×33×19=14421；         使23×28×c被33除余1，用23×28×2=1288。        因此有（5544×p+14421×e+1288×i）% lcm(23,28,33) =n+d    又23、28、33互質，即lcm(23,28,33)= 21252;       所以有n=（5544×p+14421×e+1288×i-d）%21252   本題所求的是最小整數解，避免n為負，因此最後結果為n= [n+21252]% 21252 那麼最終求解n的表達式就是：   n=(5544*p+14421*e+1288*i-d+21252)%21252;      

#include<iostream>   
#include<cstdio>   
#include<cstring>   
#include<cmath>   
#include<algorithm>   
#include<bitset>   
  
using namespace std;  
  
int main()  
{  
    int a , b , c , d ;  
    int temp ;  
    int Case ;  
    cin >> Case ;  
    while( Case-- )  
    {  
        temp = 0 ;  
        while( cin >> a >> b >> c >> d )  
        {  
            if( a == - 1 && b == -1 && c == -1 && d == -1 )  
                break;  
            int n = ( 5544 * a + 14421 * b + 1288 * c  ) % 21252 ;  
            int ans = n > d ? n - d : 21252 + n - d ;  
            cout << "Case " << ++temp << ": the next triple peak occurs in " << ans << " days." << endl;  
        }  
        if( Case != 0 )  
            cout << endl ;  
    }  
    return 0 ;  
}  

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<bitset>

using namespace std;

int main()
{
int a , b , c , d ;
int temp ;
int Case ;
cin >> Case ;
while( Case-- )
{
temp = 0 ;
while( cin >> a >> b >> c >> d )
{
if( a == - 1 && b == -1 && c == -1 && d == -1 )
break;
int n = ( 5544 * a + 14421 * b + 1288 * c  ) % 21252 ;
int ans = n > d ? n - d : 21252 + n - d ;
cout << "Case " << ++temp << ": the next triple peak occurs in " << ans << " days." << endl;
}
if( Case != 0 )
cout << endl ;
}
return 0 ;
}

 

                     

#include<iostream>   
#include<cstdio>   
#include<cstring>   
#include<cmath>   
#include<algorithm>   
#include<bitset>   
  
using namespace std;  
  
int main()  
{  
    int a , b , c , d ;  
    int temp = 0 ;  
    while( cin >> a >> b >> c >> d )  
    {  
        if( a == - 1 && b == -1 && c == -1 && d == -1 )  
            break;  
        int n = ( 5544 * a + 14421 * b + 1288 * c ) % 21252 ;  
        int ans = n > d ? n - d : 21252 + n - d ;  
        cout << "Case " << ++temp << ": the next triple peak occurs in " << ans << " days." << endl;  
    }  
    return 0 ;  
}  

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<bitset>

using namespace std;

int main()
{
int a , b , c , d ;
int temp = 0 ;
while( cin >> a >> b >> c >> d )
{
if( a == - 1 && b == -1 && c == -1 && d == -1 )
break;
int n = ( 5544 * a + 14421 * b + 1288 * c ) % 21252 ;
int ans = n > d ? n - d : 21252 + n - d ;
cout << "Case " << ++temp << ": the next triple peak occurs in " << ans << " days." << endl;
}
return 0 ;
}

 

      同樣，這道題的解法就是：    已知(n+d)%23=p;   (n+d)%28=e;   (n+d)%33=i         使33×28×a被23除余1，用33×28×8=5544；         使23×33×b被28除余1，用23×33×19=14421；         使23×28×c被33除余1，用23×28×2=1288。        因此有（5544×p+14421×e+1288×i）% lcm(23,28,33) =n+d    又23、28、33互質，即lcm(23,28,33)= 21252;       所以有n=（5544×p+14421×e+1288×i-d）%21252   本題所求的是最小整數解，避免n為負，因此最後結果為n= [n+21252]% 21252 那麼最終求解n的表達式就是：   n=(5544*p+14421*e+1288*i-d+21252)%21252;   當問題被轉化為一條數學式子時，你會發現它無比簡單。。。。直接輸出結果了。