[BZOJ2712][[Violet 2]棒球][类欧几里得算法]

[BZOJ2712][[Violet 2]棒球][类欧几里得算法]

类似于下面这道题吧，只要把小数转换成分数就好了。

http://blog.csdn.net/g1n0st/article/details/62044709

代码：

#include 
using namespace std;
typedef long long ll;
typedef pair abcd;
inline ll Min(const ll &a, const ll &b) {
    return a < b ? a : b;
}
inline ll gcd(const ll &a, const ll &b) {
    if (!b) return a;
    return gcd(b, a % b);
}
inline abcd solve(ll p1, ll q1, ll p2, ll q2) {
    ll l = p1 / q1 + 1;
    abcd ret(0,0);
    if (l * q2 < p2) return abcd(l, 1);
    if (p1 == 0) return abcd(1, q2 / p2 + 1);
    if (p1 <= q1 && p2 <= q2) {
        ret = solve(q2, p2, q1, p1);
        return abcd(ret.second, ret.first); 
    }
    ll t = p1 / q1;
    ret = solve(p1 - q1 * t, q1, p2 - q2 * t, q2);
    ret.first += ret.second * t;
    return ret;
}
int main(void) {
    //freopen("in.txt", "r", stdin);
    ll p1, q1, p2, q2, r; int n;
    while (~scanf("%d 0.%lld", &n, &r)) {
        if (r == 0) {
            printf("1\n"); continue;
        }
        ll x = 10; while (n--) x *= 10;
        p1 = r * 10 - 5, q1 = x, p2 = r * 10 + 5, q2 = x;
        ll g = gcd(p1, q1); p1 /= g, q1 /= g;
        g = gcd(p2, q2); p2 /= g, q2 /= g;
        abcd ans = solve(p1, q1, p2, q2);
        printf("%lld\n", Min(ans.second, q1));
    }
    return 0;
} 

完。

By g1n0st