Hdu3579 Hello Kiki

Hello Kiki

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4517    Accepted Submission(s): 1746

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

Author

digiter (Special Thanks echo)

Source

2010 ACM-ICPC Multi-University Training Contest（14）——Host by BJTU

分析：就是中国剩余定理的非互质版本，一定要注意余数都是0的情况.

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

using namespace std;

typedef long long ll;
const ll maxn = 100010;
ll T,n,a[maxn],m[maxn],cas;

ll gcd(ll a, ll b)
{
    if (!b)
        return a;
    return gcd(b, a % b);
}

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (!b)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll temp = exgcd(b, a % b, x, y), t = x;
    x = y;
    y = t - (a / b) * y;
    return temp;
}

ll niyuan(ll x, ll mod)
{
    ll px, py, t;
    t = exgcd(x, mod, px, py);
    if (t != 1)
        return -1;
    return (px % mod + mod) % mod;
}

bool hebing(ll a1, ll n1, ll a2, ll n2, ll &a3, ll &n3)
{
    ll d = gcd(n1, n2), c = a2 - a1;
    if (c % d != 0)
        return false;
    c = (c % n2 + n2) % n2;
    n1 /= d;
    n2 /= d;
    c /= d;
    c *= niyuan(n1, n2);
    c %= n2; //取模,在哪一个模数下就要模哪个,模数要跟着变化.
    c *= n1 * d;
    c += a1;
    n3 = n1 * n2 * d;
    a3 = (c % n3 + n3) % n3;
    return true;
}


ll solve()
{
    ll a1 = a[1],m1 = m[1],a2,m2,a3,m3;
    for (ll i = 2; i <= n; i++)
    {
        a2 = a[i],m2 = m[i];
        if (!hebing(a1,m1,a2,m2,a3,m3))
            return -1;
        a1 = a3;
        m1 = m3;
    }
    if (a1 == 0)
    {
        m1 = 1;
        for (int i = 1; i <= n; i++)
            m1 = m1 * m[i] / gcd(m1,m[i]);
        return m1;
    }
    ll t = (a1 % m1 + m1) % m1;
    return t;
}

int main()
{
    scanf("%lld",&T);
    while (T--)
    {
        scanf("%lld",&n);
        for (ll i = 1; i <= n; i++)
            scanf("%lld",&m[i]);
        for (ll i = 1; i <= n; i++)
            scanf("%lld",&a[i]);
        printf("Case %lld: %lld\n",++cas,solve());
    }

    return 0;
}