Codeforces 1244C. The Football Season

Portal

This question a $ exgcd $ 

How to look is $ exgcd $ 

Consider the $ exgcd $ how to do this problem

Consider $ wx + dy = p $ solution is obtained, and to try to meet $ x + y + z = n $

Then the solution is to ask a group of $ x, y $ $ x + y $ such as small

Because $ d <w $, so for $ wx + dy = p $

When the time taken $ Y $ of the smallest positive integer solutions $ x + y $ must be minimal

You can then be determined using the minimum $ exgcd $ $ y $, Y $ and then obtained by $ X $ $

Then $ \ text {Wrong answer on test 5} $ ....

After the end of the game found explosive $ lnog \ long $ a

The finer is required wording, when obtaining a set $ x, y $ satisfying $ wx + dy = gcd (w, d) $ after

We can not directly $ y = y * (p / gcd (w, d)) $ then $ mod = w / gcd (w, d), y = (y \% mod + mod) \% mod $

But should direct $ mod = w / gcd (w, d), y = ((y \% mod * (p / gcd (w, d)) \% mod) \% mod + mod) \% mod $

Since then $ w, d <= 10 ^ 5 $, it will not burst

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
inline ll read()
{
    ll x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') { x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }
    return x*f;
}
ll n,p,w,d;
ll exgcd(ll a,ll b,ll &x,ll &y)
{
    if(!b) { x=1,y=0; return a; }
    ll g=exgcd(b,a%b,x,y);
    ll t=x; x=y;
    y=t-a/b*y;
    return g;
}
int main()
{
    n=read(),p=read(),w=read(),d=read();
    ll x,y,g=exgcd(w,d,x,y);
    if(p%g) { printf("-1\n"); return 0; }
    ll mo=w/g;
    y=((y%mo)*((p/g)%mo)%mo+mo)%mo;
    x=(p-y*d)/w;
    if((x+y>n) || (x<0)) printf("-1\n");
    else printf("%lld %lld %lld\n",x,y,n-x-y);
    return 0;
}

exgcd practice

 

In fact because $ w, d <= 10 ^ 5 $, we can enumerate violence from small to large $ y \ in [0, w) $ and determine whether there is a legitimate $ x $

Once that is found there is a minimum of $ y $

Noting determination equation is $ p-dy = 0 \ mod w $, noted $ p-dy \ circular mod w $ segment length is also less than $ $ W

So if the enumeration finished $ y \ in [0, w) $ later still no solution then it must be no solution

Because $ w <= 10 ^ 5 $ , so to force, from the code: jiangly

#include <bits/stdc++.h>
using namespace std;
using LL = long long;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    LL n, p, d, w;
    cin >> n >> p >> d >> w;
    LL y = 0;
    while (y < d && (p - w * y) % d != 0) ++y;
    if (y < d) {
        LL x = (p - w * y) / d;
        LL z = n - x - y;
        if (x >= 0 && z >= 0) cout << x << " " << y << " " << z << "\n";
        else cout << "-1\n";
    } else cout << "-1\n";
    return 0;
}

violence