Description For sequence i1, i2, i3, … , iN, we set aj to be the number of members in the sequence which are prior to j and greater to j at the same time. The sequence a1, a2, a3, … , aN is referred to as the inversion sequence of the original sequen…

Inversion Sequence Problem's Link:   http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1555 Mean: 给你一个序列a[n],要你按照要求去构造一个序列b. 序列a[i]表示序列b中的i前面有a[i]个数比i大. 转换一下就是: 已知一个连续的序列(1,2,3,4,...),然后告诉了我们这个序列中每个数前面比本身大的个数,根据这些条件将这个序列调整顺序,找到满足条件的序列. analyse: STL大法好…

1555: Inversion Sequence Submit Page    Summary    Time Limit: 2 Sec     Memory Limit: 256 Mb     Submitted: 519     Solved: 195 Description For sequence i1, i2, i3, … , iN, we set aj to be the number of members in the sequence which are prior to j a…

1555: Inversion Sequence Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: 107  Solved: 34 Description For sequence i1, i2, i3, … , iN, we set aj to be the number of members in the sequence which are prior to j and greater to j at the same time. The seq…

1555: Inversion Sequence Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: 469  Solved: 167[Submit][Status][Web Board] Description For sequence i1, i2, i3, … , iN, we set aj to be the number of members in the sequence which are prior to j and greater to…

Inversion Sequence Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%lld & %llu Description For sequence i1, i2, i3, … , iN, we set aj to be the number of members in the sequence which are prior to j and greater to j at the same time.…

Contest2071 - 湖南多校对抗赛(2015.03.28) 本次比赛试题由湖南大学ACM校队原创 http://acm.csu.edu.cn/OnlineJudge/contest.php?cid=2071 Problem A: Rectangle Time Limit: 1 Sec  Memory Limit: 256 MBSubmit: 210  Solved: 48[Submit][Status][Web Board] Description Now ,there are some…

题意:求三个序列的最长公共子序列. 思路:一开始以为只要求出前两个的LCS,然后和第三个再求一遍LCS就是答案了.但是样例就对我进行啪啪啪打脸了.实际上就跟两个序列的差不多,换成三维的就行了. 代码:需要注意的是max速度比较慢,最后改成if #include<stdio.h> #include<string.h> #include<iostream> using namespace std; ; int dp[N][N][N]; char a[N],b[N],c[N]…

题意:给n个数,m个询问.每个询问是一个区间,求区间内差的绝对值为1的数对数. 题解:先离散化,然后莫队算法.莫队是离线算法,先按按询问左端点排序,在按右端点排序. ps:第一次写莫队,表示挺简单的,不过这题之前乱搞一气一直TLE,莫队还是很强大的. 代码: #include <algorithm> #include <iostream> #include <cstdio> #include <cmath> #include <cstring>…

莫队算法+map #include<cstdio> #include<cstring> #include<cmath> #include<map> #include<algorithm> using namespace std; +; ],pos[maxn]; struct X { int l,r,id; } s[maxn]; int L,R; int Ans,f[maxn]; bool cmp(const X&a,const X&…