题意: 给一个数X. 起始点为坐标0.第1步跳1格,第2步跳2格,第3步跳3格,.....以此类推. 每次可以向左跳或向右跳. 问最少跳几步可以到坐标X. 思路: 假设X是正数. 最快逼近X的方法是不停向右走.如果越过了X,假设到了X1,则必有X1-X小于最后一步d. 如果X1-X是偶数,将之前的某个x变为-x.则可以到X. 如果X1-X是奇数,因为将之前的某个x变为-x,实际是后退2x格.所以必定要再向右走一步.直到X1-X是偶数.(其实只要走一步或两步) X是负数的情况和X是正数的情况一样.…

B. Jumping Jack time limit per test 1 second memory limit per test 64 megabytes input standard input output standard output Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get…

Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point x. In order to train, he has decided that he'll first jump by only one unit, and each subsequent jump will be e…

Jumping Jack CodeForces - 11B 就是一个贪心. 基本思路: 正负没有关系,先取绝对值. 首先跳过头,然后考虑怎么回来. 设超过头的步数为kk.如果kk为偶数,那么直接在前面跳过来的步数中选一个kk/2的步数,改成反着跳即可,不需要额外步数. 如果kk为奇数: 显然如果只把前面跳的改成反着跳不可能导致位置与目标位置差值的奇偶性变化,而kk为奇数,需要达到的差值0是偶数,因此一定需要额外步数. 那么可能有两种情况: 设当前已经跳了p步.如果p为偶数,那么这一步的下一步(p…

B. Jumping Jack 题目连接: http://www.codeforces.com/contest/11/problem/B Description Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point x. In order to train, he has d…

B. Jumping Jack Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point x. In order to train, he has decided that he'll first jump by only one unit, and each subsequen…

NOIP2012BLOCKADE贪心思想证明 这道题的做法是二分时间并检验这个时间是否可行.检验的方法要用到贪心思想. 对于不能到根结点的军队应该尽量向根结点走. 如果军队A能走到根结点但到根结点后剩余的时间不够返回根结点的儿子B,应该让军队A守B. 否则把军队A加入优先队列,用剩余时间小的军队匹配到根结点距离小的儿子. 证明: 一个军队可控制自己和以自己为根的子树.越向上走控制的结点越多. 把这个军队记作A,如果不这样,我们必须要用另一个到达根结点的军队,记作C,因为A剩余时间小于到B的时间,…

题目传送门 /* 贪心/数学:还以为是BFS,其实x1 + 4 * k = x2, y1 + 4 * l = y2 */ #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; const int INF = 0x3f3f3f3f; char s[MAXN][MAXN]; int main(void) //Codeforces Round #212 (Div. 2)…

[HNOI2015]菜肴制作贪心的证明 先吐槽一句为什么网上都没人证这个东西,我觉得一点也不显然啊... 判环不用说了,现在处理一个DAG.考虑按题意模拟:建反图(边从后选的点连向先选的点),每次找全局最小的点,并把候选集合缩小为这个点在图上的所有后继,再在当前候选集合中选择最小的点并把候选集合更新为它的后继,不断重复此过程直至候选集合只有一个点,就把这个点从图上删除并放到答案序列的末尾.这种做法最后得到的答案序列的正确性是显然的. 观察一下数据范围,发现直接模拟不太能做.考虑每次选了全局最小点…

题意: m个水果,n个价格.每种水果只有一个价格. 问如果给每种水果分配价格,使得买的m个水果总价格最小.最大. 输出最小值和最大值. 思路: 贪心. 代码: bool cmp(int a,int b){ return a>b; } string name; map<string,int> mp; int price[200],fruit[200]; int cn; int n,m; int main(){ cin>>n>>m; mp.clear(); cn=0;…