CF1244C The Football Season

题目链接

problem

给定\(n,p,w,d\)，求解任意一对\((x,y)\)满足\[xw+yd=p\\ x + y \le n\]

\(1\le n\le 10^{12},0\le p\le 10^{17},1\le d<w \le 10^5\)

solution

注意到\(n,p\)非常大，\(w,d\)比较小。而且\(w>d\)。所以我们就想让\(y\)尽量小。

实际上如果最终有解，那在\(y\le w\)中肯定有解。

证明如下:

如果有\(y'=aw+k(a\ge 1,0\le k < w)\)使得\(xw+y'd=p\)。即\(xw+(aw+k)d=xw+awd+kd=(x+ad)w+kd=p\)。

发现\(xw+(aw+k)d\)的系数和为\(x+aw+k\)。\((x+ad)w+kd\)的系数和为\(x+ad+k\)。又因为\(w>d\)。所以后者的系数和要小。所以\(d\)的系数一定小于等于\(w\)

然后在区间\([0,w]\)中枚举\(y\)。计算\(x\)即可。

code

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
#include<cmath>
#include<map>
#include<string>
using namespace std;
typedef long long ll;

ll read() {
    ll x = 0,f = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0',c = getchar();}
    return x * f;
}

int main() {
    ll n = read(),p = read(),w = read(),d = read();

    for(ll y = 0;y <= w;++y) {
        if((p - y * d) % w) continue;
        ll x = (p - y * d) / w;
        if(x >= 0 && x + y <= n) {
            printf("%i64d %i64d %i64d\n",x,y,n - x - y);
            return 0;
        }
    }
    puts("-1");
    return 0;
}