蛤蛤,略蠢. priority_queue 自定义优先级 和排序是反的 struct node { int x,y; friend bool operator< (node a,node b) { if(a.y<b.y) return 1; if(a.y==b.y&&a.x>b.x) return 1; return 0; } }; priority_queue<node>qu; 思路: +v就加一下,标记一下 对了-v 遍历一下,标记一下 加个优先队列维护一…

1014 - Absolute Defeat Time Limit:2s Memory Limit:64MByte Submissions:257Solved:73 DESCRIPTION Eric has an array of integers a1,a2,...,ana1,a2,...,an. Every time, he can choose a contiguous subsequence of length kk and increase every integer in the c…

1152 - Expected value of the expression Time Limit:2s Memory Limit:128MByte Submissions:128Solved:63 DESCRIPTION You are given an expression: A0O1A1O2A2⋯OnAnA0O1A1O2A2⋯OnAn, where Ai(0≤i≤n)Ai(0≤i≤n) represents number, Oi(1≤i≤n)Oi(1≤i≤n) represents op…

分析:a^b+2(a&b)=a+b  so->a^(-b)+2(a&(-b))=a-b 然后树状数组分类讨论即可 链接:http://www.ifrog.cc/acm/problem/1023 吐槽:这个题本来是mod(2^40),明显要用快速乘啊,但是用了以后狂T,不用反而过了,不懂出题人 #include <iostream> #include <algorithm> #include <cmath> #include <vector&g…

http://www.ifrog.cc/acm/problem/1028 题解处:http://www.ifrog.cc/acm/solution/4 #include <cstdio> #include <cstring> #include <queue> #include <cmath> #include <algorithm> using namespace std; typedef long long LL; ; ; const doub…

1010 - Alarm Time Limit:1s Memory Limit:128MByte DESCRIPTION Given a number sequence [3,7,22,45,116,...] . Please tell me the k -th number. INPUT A number T (T<100) indicates the number of the input cases. Then for each case there only is one integer…

1050 - array Time Limit:3s Memory Limit:64MByte Submissions:494Solved:155 DESCRIPTION 2 array is an array, which looks like: 1,2,4,8,16,32,64......a1=1 ,a[i+1]/a[i]=2; Give you a number array, and your mission is to get the number of subsequences ,wh…

1052 - See car Time Limit:2s Memory Limit:64MByte Submissions:594Solved:227 DESCRIPTION You are the god of cars, standing at (a, b) point.There are some cars at point (xi,yi), . If lots of cars and you are in one line, you can only see the car that i…

SAMPLE INPUT 3 2 2 2 1 1 5 1 4 1 2 3 4 5 4 10 3 1 2 3 4 SAMPLE OUTPUT 1 0 15 前缀和,每个元素都判断一下. #include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <cmath> #include <map> #include <set>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1072 思路:深搜每一个节点,并且进行剪枝,记录每一步上一次的s1,s2:如果之前走过的时间小于这一次, 就说明有更短的:路径,所以就不用继续遍历下去. #include<iostream> #include<cstdio> #include<cstring> using namespace std; ][],step[][],tim[][],m,n,ans; ][]={{-…