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 ;

}