uva 10090 - Marbles(欧几里得+通解)

题目链接:uva 10090 - Marbles

题目大意:给出n,表示有n个珠子,现在要用若干个盒子来装。有两种盒子,一种价钱c1,可以装t1个珠子,另一种价钱c2,可以装t2个珠子。要求所卖的盒子刚好装n个珠子,并且价钱最小的方案。

解题思路:用拓展欧几里得算法求出xt1+yt2=n的一对解xy,这样就有通解:
x=xngcd(t1,t2)+t2gcd(t1,t2)k
y=yngcd(t1,t2)t1gcd(t1,t2)k

然后根据性价比选择一种盒子的个数尽量多。

#include <cstdio>
#include <cstring>
#include <cmath>

typedef long long ll;

ll n, c1, t1, c2, t2;

void gcd (ll a, ll b, ll& d, ll& x, ll& y) {
    if (!b) {
        d = a;
        x = 1;
        y = 0;
    } else {
        gcd(b, a%b, d, y, x);
        y -= (a/b)*x;
    }
}

int main () {
    while (scanf("%lld", &n) == 1 && n) {
        scanf("%lld%lld%lld%lld", &c1, &t1, &c2, &t2);
        ll d, xi, yi, x, y;
        gcd(t1, t2, d, xi, yi);
        ll lower = ceil(-1.0 * n * xi / t2);
        ll up = floor(1.0 * n * yi / t1);

        if (lower > up || n % d) {
            printf("failed\n");
            continue;
        }

        if (c1 * t2 > c2 * t1) {
            x = xi * n / d + t2 / d * lower;
            y = yi * n / d - t1 / d * lower;
        } else {
            x = xi * n / d + t2 / d * up;
            y = yi * n / d - t1 / d * up;
        }
        printf("%lld %lld\n", x, y);
    }
    return 0;
}

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