hdu 5109 同余定理＋枚举＋构造法求解

来源：互联网发布：linux安装多个jdk版本编辑：程序博客网时间：2024/06/06 03:34

因为S是T的子串，那么我们可以构建数XSY,那么这个数就是num = (X*10^strlen(S) + S )*10*strlen(Y) + Y，因为num%a == 0 ，所以根据同余模定理，

num%a == ((x%a) * (( 10^strlen(S) + S%a) *10*strlen(Y) )%a + Y%a,

设　mod = ( num - Y%a ) %a；

所以　Y%a = ( a - mod ) %a;

所以只需枚举Y的长度和x%a，此问题得解

#include <cstdio>#include <algorithm>#include <cstring>#include <iostream>using namespace std;typedef long long LL;int main ( ){    LL a , s, b;    char str[10];    while ( ~scanf ( "%lld" , &a ) )    {        scanf ( "%s" , &str );        LL base = 1;        int len = strlen ( str );        if ( len == 1 && str[0] =='0' )         {            puts ( "0" );            continue;        }        for ( int i = 0 ; i < len ; i++ )            base *= 10L;        sscanf ( str , "%I64d" , &s );        b = -1;        for ( LL i = 1 ; i <= 10000 ; i *= 10 )            for ( LL j = (str[0]=='0'?1:0);j < a ; j++ )            {                LL temp = ( j*base + s ) * i;                LL y = ( a - temp%a ) %a;                if ( y  < i )                {                    temp += y;                    if ( b == -1 || temp < b )                        b = temp;                }             }        printf ( "%I64d\n" , b/a);     }}

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