尺取法 poj 2566】的更多相关文章

尺取法:顾名思义就是像尺子一样一段一段去取,保存每次的选取区间的左右端点.然后一直推进 解决问题的思路: 先移动右端点 ,右端点推进的时候一般是加 然后推进左端点,左端点一般是减 poj 2566 题意:从数列中找出连续序列,使得和的绝对值与目标数之差最小 思路: 在原来的数列开头添加一个0 每次找到的区间为 [min(i,j)+1,max(i,j)] 应用尺取法的代码: while (r <=n) { int sub = pre[r].sum - pre[l].sum; while (abs(…
题目传送门 /* 尺取法:先求出不同知识点的总个数tot,然后以获得知识点的个数作为界限, 更新最小值 */ #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <set> #include <map> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MA…
题目传送门 /* 题意:求连续子序列的和不小于s的长度的最小值 尺取法:对数组保存一组下标(起点,终点),使用两端点得到答案 1. 记录前i项的总和,求[i, p)长度的最小值,用二分找到sum[p] - s[i] >= s的p 2. 除了O (nlogn)的方法,还可以在O (n)实现,[i, j)的区间求和,移动两端点,更新最小值,真的像尺取虫在爬:) */ #include <cstdio> #include <algorithm> #include <cstri…
Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13955   Accepted: 5896 Description A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) ar…
给一个数 写成连续质数的和的形式,能写出多少种 *解法:先筛质数 然后尺取法 **尺取法:固定区间左.右端点为0,如果区间和比目标值大则右移左端点,比目标值小则右移右端点                 详见http://blog.csdn.net/consciousman/article/details/52348439 #include <iostream> #include <cstdio> using namespace std; #define SZ 11000 bool…
Description Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration (that must be going through a defiant phase: "But I want to use feet, not meters!"). Each signal seems to…
传送门 参考资料: [1]:http://www.voidcn.com/article/p-huucvank-dv.html 题意: 题意就是找一个连续的子区间,使它的和的绝对值最接近target. 题解: 这题的做法是先预处理出前缀和,然后对前缀和进行排序,再用尺取法贪心的去找最合适的区间. 要注意的是尺取法时首尾指针一定不能相同,因为这时区间相减结果为0,但实际上区间为空,这是不存在的,可能会产生错误的结果. 处理时,把(0,0)这个点也放进数组一起排序,比单独判断起点为1的区间更方便. 还…
Bound Found Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 5207   Accepted: 1667   Special Judge Description Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration…
Bound Found Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 2277   Accepted: 703   Special Judge Description Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration…
poj3061 Subsequence 题目链接: http://poj.org/problem?id=3061 挑战P146.题意:给定长度为n的数列整数a0,a1,...,a(n-1)以及整数S,求出总和不小于S的连续子序列的长度的最小值,如果解不存在,则输出零.$10<n<10^5,0<a_i<=10^4,S<10^8$; 思路:尺取法,设起始下标s,截止下标e,和为sum,初始时s=e=0;若sum<S,将sum增加a(e),并将e加1,若sum>=S,更…