POJ 1006 同余方程组

以前的做法

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
LL ai[],ri[],M;
void Exgcd(LL a,LL b,LL& d,LL& x,LL& y)
{
    if(b == ) { d = a, x = , y = ;}
    else {
        Exgcd(b,a%b,d,y,x); y -= x*(a / b);
    }
}
LL gcd(LL a,LL b)
{
    return b ==  ? a : gcd(b,a%b);
}
LL solve(int n)
{
    LL a1,r1;
    LL a, b, c;
    LL x, y, d;
    a1 = ai[], r1 = ri[];
    for(int i = ; i <= n; ++i)
    {
        a = a1, b = ai[i], c = ri[i] - r1;
        Exgcd(a,b,d,x,y);
        if(c % d) return -;
        LL t = b / d;
        x = (x * (c / d)% t + t) % t;
        r1 = a1 * x + r1;
        a1 = a1 *(ai[i] / d);
    }
    return r1;
}
int main()
{
    LL p,e,k,d;
    int Kase = ;
    while(cin >> p >> e >>k >>d)
    {
        ai[] = , ai[] = , ai[] = ;
        ri[] = p , ri[] = e, ri[] = k;
        if(p == - && e == - && k == - && d == -) break;
        LL ans = solve();
        M = ;
        for(int i = ; i <= ; ++i) M = M * ai[i] / gcd(M,ai[i]);
        if(ans >= M) ans %= M; // 求最小的值.
        while(ans <= d) ans += M;
        printf("Case %d: the next triple peak occurs in %lld days.\n",Kase++,ans-d);
    }
}

中国剩余定理撸一发

不过中国剩余定理 ai[i] 需要互质

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
LL ai[],ri[],M;
void Exgcd(LL a,LL b,LL& d,LL& x,LL& y)
{
    if(b == ) { d = a, x = , y = ;}
    else {
        Exgcd(b,a%b,d,y,x); y -= x*(a / b);
    }
}
LL gcd(LL a,LL b)
{
    return b ==  ? a : gcd(b,a%b);
}
LL China(int n)
{
    M = ;
    LL Mi,x,y,d,ans = ;
    for(int i = ; i <= n; ++i) M *= ai[i];
    for(int i = ; i <= n; ++i)
    {
        Mi = M / ai[i];
        Exgcd(Mi,ai[i],d,x,y);
        ans = (ans + Mi * x * ri[i]) % M;
    }
    if(ans < ) ans += M;
    return ans ;
}
int main()
{
    LL p,e,k,d;
    int Kase = ;
    while(cin >> p >> e >>k >>d)
    {
        ai[] = , ai[] = , ai[] = ;
        ri[] = p , ri[] = e, ri[] = k;
        if(p == - && e == - && k == - && d == -) break;
        LL ans = China();
        /*M = 1;
        for(int i = 1; i <= 3; ++i) M = M * ai[i] / gcd(M,ai[i]);*/
        //if(ans >= M) ans %= M; // 求最小的值.
        while(ans <= d) ans += M;
        printf("Case %d: the next triple peak occurs in %lld days.\n",Kase++,ans-d);
    }
}