title: woj1010 alternate sum 数学
date: 2020-03-10
categories: acm
tags: [acm,woj,数学]
一道数学题。简单。
题意
给一个集合Q，对于每个子集中元素非降序排列后，计算a1-a2+a3-a4....=alternate sum。
比如 Q={2,1,3,4}，其本身alternate sum=4-3+2-1=2，而其中一个子集｛1，3，4｝:4-3+1=2;
求Q中所有子集的alternate sum之和。
输入格式
There are several test cases. For each test case, it contains:
Line 1: One integer N (1<=N<=1000) which specifies the size of set Q
Line 2: N integers (you are ensured that the absolute value of each integer is less than 2^16) which specify the elements of set Q.
There is a blank between two consecutive integers.
The input will be ended by zero.
输出格式
The answer of the total value may be very large, so you should divide this answer by 2006 and just output the remainder instead.
样例输入
1
-1
2
0 4
3
1 3 2
0
样例输出
2005
8
12
分析:这种一看是数学题，就在纸上算算。
一共有2^N个子集,单个计算是不现实的。
对于集合中最大元素maxn，包含它的子集有2^(N-1)个。其在计算中必然为 + 。有2^(N-1)个
对于第二大元素a2:包含它的子集一共有2^(N-1)个，其中一半有maxn，在计算时-a2，一半+a2。所以总的下来没影响。
其他数以此类推。
所以最后结果为maxn*2(N-1)
要说怎么想到的：学集合的时候知道子集数量是2N，其中2就相当于每个数取不取，然后就想到2(N-1)...
code
#include<iostream>
#include<cstdio>
using namespace std;
int maxn,n,ans,num,i;
int main(){
while(scanf("%d",&num)==1&&num){
cin>>n;
maxn=n;
for(i=0;i<num-1;i++){
cin>>n;
maxn=(maxn>n)?maxn:n;
}
maxn=(maxn%2006+2006)%2006; //保证是正数
for(i=0;i<num-1;i++){
maxn=maxn*2;
maxn=maxn%2006; //防溢出
}
cout<<maxn<<endl;
}
return 0;
}
被除数÷除数=商……余数。
由此可知，余数=被除数-商×除数 (*)
以下摘录自C++ Primer（P130）
操作符%称为“求余”或“求模”操作符，该操作符的操作数只能为整型。
如果两个操作数为正，结果也为正；如果两个操作数都为负数，结果也为负数；如果一个操作数为正数，一个操作数为负数，求模结果的符号取决于机器。
当操作数中有一个为负，一个为正是，求模操作结果值的符号可依据分子（被除数）或分母（除数）的符号而定。
如果求模的结果随分子的符号，则除出来的值向零一侧取整；如果求模与分母的符号匹配，则除出来的值向负无穷大一侧取整。
取模运算https://baike.baidu.com/item/%E5%8F%96%E6%A8%A1%E8%BF%90%E7%AE%97/10739384?fr=aladdin
title: woj1011 Finding Teamates 数学
date: 2020-03-12
categories: acm
tags: [acm,数学,woj]
一般题。数学(思维).
description
Big day is coming!
The Wuhan University Collegiate Programming Contest (WHUCPC 2006) will be held in this April. All the ACM lovers, excellent programmers and
computer geniuses in Wuhan University are of great joy, sweeping lots of problems at the Online Judges as preparation, waiting for this big day
to come.However, some students are still in trouble, because they have not found their teammates yet. So the WHUCPC Committee decided to
help these students. All the students that have no teammates are gathered in Peng Kun Square. The method to help them to find their
teammates is quite straightforward: all the students stand in a row, if some consecutive students form an increasing sequence
in their height, then these students must be put in the same team. That is, if there are 4 students who form an increasing sequence,
they must be put in one team and can't be divided long longo more than one team.
For example, if there are six students, standing as the following figure shows:
1 3 4 5 2 6
Where 1 represents the shortest student, and 6 represents the tallest student, etc.
Since (1, 3, 4, 5) forms an increasing sequence, they should put in the same team, and 2, 6 forms another increasing sequence,
so 2, 6 are put long longo another team. However, as we all know, the ACM teams are required 3 members each. In the above case, team 1
has 4 members, so the above alignment is invalid. The following figure shows some of the valid alignments:
1 3 4 2 5 6
But in some special cases, if the number of students cannot be divided evenly by 3, we ALLOW the LAST team has less than 3 members.
So if there are 5 students, the follow figure shows some of the valid and invalid alignments:
2 3 4 1 5 (ok) {234} {15}
1 5 2 3 4 (no) {15} {2 3 4 } team1 is not the last team
Now, given the number of the students, your mission is to tell us how many valid alignments there are.
Note, no two students are of the same height.
input
There are several test cases.
Each case contains a single long longeger n, representing the number of students who need to find teammates. (1<=n<=25)
output
Output an long longeger in a single line, which is the number of valid alignments. You are ensured that the result is less than 2^63.
Sample input
2
3
4
6
9
Sample output
1
1
3
19
1513
analyse
3人一队，n形如3x+y，y=012.
一看感觉是递推(直接从n想我是想不出来)，先想了catalan，然后不对，估计就是自己推了。
假设知道了f(n)，然后加了一个数y，那么至少有一种解就是y放在最后，ans+f(N)
然后y放到前面某个位置，就发现，其实相当于把n+1序列分成两个子序列（因为是递增无重复）f（x）f（n+1-x）。
这里每一组都可以看成1...x，因为无重复且递增，就有了递推
然后考虑y的位置，假设index从1开始，那必须是3x，否则不满足3人一队的条件。这样选择有n/3种。
然后这n/3种位置就有n/3种f（3x）f（n+1-3x）。
设x为数组下标，那么设y到了x组，这种方案，为f(3x-1)(y固定在3x位置)f(n+1-3x)(后面的)*C(n,3x-1) (挑出放在前面的3x-1个数)
最后想特殊情况，没想到，就代入4试试
带入4算得4，这时候想到问题是，1234这种不符合要求(刚刚忘了排除了)
解决，直接n+1/3找合适位置
然后又有问题是当n+1%2==2时其实是可以放在最后的，特判就行
然后就这样写吧
最后是经验：想不出来，就试试递推
先想一般情况
再考虑有没有特殊情况
代入试一试
code
#include<iostream>
#include<cstdio>
using namespace std;
long long combine(int n,int m){
if(m>n-m) //一开始没有优化错了。一方面减循环，一方面减少乘法溢出
m=n-m;
long long x=1,y=1,z=2;
for(int i=n-m+1;i<=n;i++)
x*=i;
for(int i=1;i<=m;i++)
y*=i;
return x/y;
}
long long func(int n){
/*
if(n==0)
return 0; //一开始写成0了....这样返回后一乘就没了....
*/
if(n>=0&&n<4)
return 1;
int cnt=(n)/3;
long long ans=0; //注意ll
for(int i=1;i<=cnt;i++){
ans+=func(3*i-1)*func(n-3*i)*combine(n-1,3*i-1);
}
if(n%3==2)
ans+=func(n-1);
return ans;
}
int main(){
int n;
while(cin>>n){
cout<<func(n)<<endl;
}
return 0;
}