hdu 3579(中国剩余定理+考虑0)

Hello Kiki

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3126    Accepted Submission(s): 1164

Problem Description

One
day I was shopping in the supermarket. There was a cashier counting
coins seriously when a little kid running and singing "门前大桥下游过一群鸭，快来快来
数一数，二四六七八". And then the cashier put the counted coins back morosely and
count again...
Hello Kiki is such a lovely girl that she loves doing
counting in a different way. For example, when she is counting X coins,
she count them N times. Each time she divide the coins into several
same sized groups and write down the group size Mi and the number of the
remaining coins Ai on her note.
One day Kiki's father found her note and he wanted to know how much coins Kiki was counting.

 

Input

The first line is T indicating the number of test cases.
Each
case contains N on the first line, Mi(1 <= i <= N) on the second
line, and corresponding Ai(1 <= i <= N) on the third line.
All numbers in the input and output are integers.
1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi

 

Output

For
each case output the least positive integer X which Kiki was counting
in the sample output format. If there is no solution then output -1.

 

Sample Input

2
2
14 57
5 56
5
19 54 40 24 80
11 2 36 20 76

 

Sample Output

Case 1: 341
Case 2: 5996

 

 

output the least positive integer X

注意这句话，当为0的时候输出mod

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <math.h>
using namespace std;
typedef long long LL;
const int N = ;
LL extend_gcd(LL a,LL b,LL &x,LL &y)
{
    if(b==)
    {
        x=,y=;
        return a;
    }
    else
    {
        LL x1,y1;
        LL d = extend_gcd(b,a%b,x1,y1);
        x = y1;
        y = x1-a/b*y1;
        return d;
    }
}
LL m[N],a[N];///模数为m,余数为a, X % m = a
bool solve(LL &m0,LL &a0,LL m,LL a)
{
    long long y,x;
    LL g = extend_gcd(m0,m,x,y);
    LL t = a-a0>?a-a0:a0-a;
    if( t%g )return false;
    x *= (a - a0)/g;
    x %= m/g;
    a0 = (x*m0 + a0);
    m0 *= m/g;
    a0 %= m0;
    if( a0 <  )a0 += m0;
    return true;
}
/**
* 无解返回false,有解返回true;
* 解的形式最后为 a0 + m0 * t (0<=a0<m0)
*/
bool MLES(LL &m0 ,LL &a0,LL n)///解为 X = a0 + m0 * k
{
    bool flag = true;
    m0 = ;
    a0 = ;
    for(int i = ; i < n; i++)
        if( !solve(m0,a0,m[i],a[i]) )
        {
            flag = false;
            break;
        }
    return flag;
}
int main()
{
    int n;
    int tcase;
    scanf("%d",&tcase);
    int t=;
    while(tcase--)
    {
        scanf("%d",&n);
        for(int i=; i<n; i++)
        {
            scanf("%lld",&m[i]);
        }
        for(int i=; i<n; i++)
        {
            scanf("%lld",&a[i]);
        }
        LL m0,a0;
        bool flag = MLES(m0,a0,n);
        LL x = (a0%m0+m0)%m0;
        printf("Case %d: ",t++);
        if(!flag) printf("-1\n");
        else
        {
            if(x==) x+=m0;
            printf("%lld\n",x);
        }
    }
    return ;
}