hdu5338 ZZX and Permutations 非原创,来自多校题解 不是自己写的,惭愧ing…… 留着以后自己参考…… lower_bound {1,2,4,5} 询问 2,返回的是 2 ,询问3 返回的是 4 是大于等于元素的值 upper_bound {1,2,4,5} 询问2,返回4,询问3,返回4,是 大于 元素的值 题意:图论的知识 1 2 3 4 5 6 (1) 1 4 5 6 3 2 (2) (1)中 数字 1 的 位置没变 所以(1) 2 的为位置 编程了 4 ,4…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud ZZX and Permutations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 181    Accepted Submission(s): 38 Problem Description ZZX likes…

ZZX and Permutations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 24    Accepted Submission(s): 2 Problem Description ZZX likes permutations. ZZX knows that a permutation can be decomposed…

题目传送门 /* 题意:不懂... 线段树+树状数组+贪心:贪心从第一位开始枚举,一个数可以是循环节的末尾或者在循环节中,循环节(循环节内部是后面的换到前面,最前面的换到最后面).线段树维护最大值,树状数组维护区间是否是循环节,查找前面最左边不是循环节的可用二分.我还是云里雾里的,看懂了网上的解题报告但还是不是完全明白题意:( 详细解释:http://blog.csdn.net/qq_24451605/article/details/47173933 */ /******************…

pid=5338" target="_blank" style="text-decoration:none; color:rgb(45,125,94); background-color:transparent">链接 多校题解 胡搞.. . 题意太难懂了. . ZZX and Permutations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (J…

ZZX and Permutations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 771    Accepted Submission(s): 243 Problem Description ZZX likes permutations. ZZX knows that a permutation can be decompos…

ZZX and Permutations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 888    Accepted Submission(s): 278 Problem Description ZZX likes permutations. ZZX knows that a permutation can be decompo…

1001 Olympiad 签到题1. # include <iostream> # include <cstdio> using namespace std; ]={}; bool judge(int x) { ]={}; ) { cnt[x%]++; ]>) return false; x/=; } return true; } int main(void) { ;i<=;i++) ]+; ]; int T; cin>>T; while(T--) { i…

Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example,[1,1,2] have the following unique permutations: [ [1,1,2], [1,2,1], [2,1,1] ] 分析: 全组合的思想,保证start和end之间交换的时候中间没有与end相同的数字 class Solution…

Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example,[1,1,2] have the following unique permutations:[1,1,2], [1,2,1], and [2,1,1]. 这道题是之前那道Permutations 全排列的延伸,由于输入数组有可能出现重复数字,如果按照之前的算法运算,会有…