Foj1675数论

http://acm.fzu.edu.cn/problem.php?pid=1675

首先必须知道一点数论的基本公式 (a-b) %c =0 -----> a%c=b%c

首先通过大数取余求出目标数值对77的余数，然后求出要求数值之后开始到最后的数值对于77取余得到的余数，满足条件然后输出结果；进行枚举就可以了。

 

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

using namespace std;
double num[]={1.0,10.0,100.0,1000.0,10000.0,100000.0,1000000.0,10000000.0,100000000.0,1000000000.0,10000000000.0,
100000000000.0,1000000000000.0,10000000000000.0,100000000000000.0,1000000000000000.0};
 
int main()
{
	__int64 i , j , k , x , y , len , temp , sum , sum1 , sum2 , sum3 ;
	char str[ 1000005 ] ;
	while( scanf( "%s" , str ) != EOF )
	{
		len = strlen( str ) ;
		sum = sum2 = 0 ;
		for( i = 0 ; str[ i ] ; ++i )
		{
			if( str[ i ] == 'x' )
			{ 
				sum = ( sum * 100 + 99 ) % 77 ;
				break ;
			}
			sum = sum * 10 + ( str[ i ] - '0' ) ;
			sum %= 77 ;
			printf( "%c" , str[ i ] ) ;
		}
		for( j = i + 4 ; str[ j ] ; ++j )
		{
			sum2 = sum2 * 10 + ( str[ j ] - '0' ) ;
			sum2 %= 77 ;
		}
		temp = len - i - 4 ;
		for( k = 23 ; k <= 99 ; ++k )
		{
			sum1 = ( sum * 100 + k ) % 77 ;
			for( x = 0 ; x <= temp / 15 + 1 ; ++x )
			{
				sum1 = sum1 * num[ 15 ] ;
				sum1 %= 77 ;
			}
			sum1 = sum1 * num[ temp % 15 ] + sum2 ;
			sum1 %= 77 ;
			if( sum1 == 0 )
				break ;
		}
		printf( "99%d" , k ) ;
		for( y = i + 4 ; str[ y ] ; y++ )
			printf( "%c" , str[ y ] ) ;
		printf( "\n" );
	}
	return 0 ;
}