SGU141 Jumping Joe

题目大意

有只青蛙在x轴原点，要刚好K步跳到坐标为P的点上
每次可以向左或向右跳x1或x2的距离，构造跳的方式，不存在则输出“NO”

算法思路

对于同样的跳跃距离，向左跳与向右跳互相抵消，因此每种跳跃只会朝一个方向
假设跳了a次x1，跳了b次x2，正值向右，负值向左
则该问题为求解方程 a * x1 + b * x2 = P，考察参数t满足 | t | <= | P | + 1 的所有解
求出a、b后，考察a、b绝对值之和与K是否相差为偶数，是的话剩下的步数，可以一左一右跳完

时间复杂度: O(P)

代码

/** * Copyright © 2015 Authors. All rights reserved. *  * FileName: 141.cpp * Author: Beiyu Li <[e-mail]> * Date: 2015-06-17 */#include <bits/stdc++.h>using namespace std;#define rep(i,n) for (int i = 0; i < (n); ++i)#define For(i,s,t) for (int i = (s); i <= (t); ++i)#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)typedef long long LL;typedef pair<int, int> Pii;const int inf = 0x3f3f3f3f;const LL infLL = 0x3f3f3f3f3f3f3f3fLL;void gcd(LL a, LL b, LL &g, LL &x0, LL &y0){        if (!b) g = a, x0 = 1, y0 = 0;        else gcd(b, a % b, g, y0, x0), y0 -= x0 * (a / b);}LL a, b, c, k;bool calc(LL x, LL y){        LL t = abs(x) + abs(y);        if (t > k || ((k - t) & 1)) return false;        puts("YES");        k = (k - t) / 2;        if (x >= 0) printf("%lld %lld ", x + k, k);        else printf("%lld %lld ", k, -x + k);        if (y >= 0) printf("%lld %lld\n", y, 0LL);        else printf("%lld %lld\n", 0LL, -y);        return true;}bool solve(){        LL g, x0, y0;        gcd(a, b, g, x0, y0);        if (c % g) return false;        a /= g, b /= g, c /= g;        For(t,-abs(c)-1,abs(c)+1)                if (calc(x0 * c + b * t, y0 * c - a * t)) return true;        return false;}int main(){        scanf("%lld%lld%lld%lld", &a, &b, &c, &k);        if (!solve()) puts("NO");        return 0;}

---