Unique Encryption Keys 题目链接: http://acm.hust.edu.cn/vjudge/problem/26633 Description http://7xjob4.com1.z0.glb.clouddn.com/bced90d15c27e75270cc759651dcfa63 Input The input contains several cipher descriptions. Each description starts with one line co…

DES:给出一个数列.然后有q个询问,对给定的区间内是否有重复的数.如果有输出任意一个重复的.如果没有输出OK. 开始以为是线段树.后来发现.只是水的比较隐蔽.感觉这个方法还是很聪明的.把每个点的最近的有重复的数的区间记录下来.这样每次询问都可以直接表示结果了. #include<stdio.h> #include<string.h> #include<iostream> #include<map> #define inf 0x3f3f3f3f using…

UVALive - 4108 SKYLINE Time Limit: 3000MS     64bit IO Format: %lld & %llu Submit Status uDebug Description   The skyline of Singapore as viewed from the Marina Promenade (shown on the left) is one of the iconic scenes of Singapore. Country X would a…

UVALive - 3942 Remember the Word A potentiometer, or potmeter for short, is an electronic device with a variable electric resistance. It has two terminals and some kind of control mechanism (often a dial, a wheel or a slide) with which the resistance…

UVALive - 3942 Remember the Word Neal is very curious about combinatorial problems, and now here comes a problem about words. Know- ing that Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie. Since Jiejie can’t remem…

题目传送门 /* 题意:本来有n个雕塑,等间距的分布在圆周上,现在多了m个雕塑,问一共要移动多少距离: 思维题:认为一个雕塑不动,视为坐标0,其他点向最近的点移动,四舍五入判断,比例最后乘会10000即为距离: 详细解释:http://www.cnblogs.com/zywscq/p/4268556.html */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath&…

题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4156 题目拷贝难度大我就不复制了. 题目大意:维护一个字符串,要求支持插入.删除操作,还有输出第 i 次操作后的某个子串.强制在线. 思路1:使用可持久化treap可破,详细可见CLJ的<可持久化数据结构的研究>. 思路2:rope大法好,详见:http…

Permutation Graphs Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVALive 6508 #include<stdio.h> #include<string.h> ],a[],b[],c[],b1[]; long long num; void merg_sort(int a[],int l,int r) { int…

Boxes Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVALive 6500 #include<stdio.h> #include<string.h> int main() { int T,m,n; int i,j,s; ][]; scanf("%d",&T); while(T--) { s=; sc…

题目链接:UVALive 6948  Jokewithpermutation 题意:给一串数字序列,没有空格,拆成从1到N的连续数列. dfs. 可以计算出N的值,也可以直接检验当前数组是否合法. #include <stdio.h> #include <iostream> #include <string.h> #define maxn 100 using namespace std; char str[maxn]; int num[maxn]; bool vis[m…