题目背景
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题目描述
The N (1 <= N <= 20) cows conveniently numbered 1...N are playing yet another one of their crazy games with Farmer John. The cows will arrange themselves in a line and ask Farmer John what their line number is. In return, Farmer John can give them a line number and the cows must rearrange themselves into that line.
A line number is assigned by numbering all the permutations of the line in lexicographic order.
Consider this example:
Farmer John has 5 cows and gives them the line number of 3.
The permutations of the line in ascending lexicographic order: 1st: 1 2 3 4 5
2nd: 1 2 3 5 4
3rd: 1 2 4 3 5
Therefore, the cows will line themselves in the cow line 1 2 4 3 5.
The cows, in return, line themselves in the configuration ‘1 2 5 3 4‘ and ask Farmer John what their line number is.
Continuing with the list:
4th : 1 2 4 5 3
5th : 1 2 5 3 4
Farmer John can see the answer here is 5
Farmer John and the cows would like your help to play their game. They have K (1 <= K <= 10,000) queries that they need help with. Query i has two parts: C_i will be the command, which is either ‘P‘ or ‘Q‘.
If C_i is ‘P‘, then the second part of the query will be one integer A_i (1 <= A_i <= N!), which is a line number. This is Farmer John challenging the cows to line up in the correct cow line.
If C_i is ‘Q‘, then the second part of the query will be N distinct integers B_ij (1 <= B_ij <= N). This will denote a cow line. These are the cows challenging Farmer John to find their line number.
N(1<=N<=20)头牛，编号为1...N，正在与FJ玩一个疯狂的游戏。奶牛会排成一行（牛线），问FJ此时的行号是多少。之后，FJ会给牛一个行号，牛必须按照新行号排列成线。
行号是通过以字典序对行的所有排列进行编号来分配的。比如说：FJ有5头牛，让他们排为行号3，排列顺序为：
1：1 2 3 4 5
2：1 2 3 5 4
3：1 2 4 3 5
因此，牛将在牛线1 2 4 3 5中。
之后，奶牛排列为“1 2 5 3 4”，并向FJ问他们的行号。继续列表：
4：1 2 4 5 3
5：1 2 5 3 4
FJ可以看到这里的答案是5。
FJ和奶牛希望你的帮助玩他们的游戏。他们需要K(1<=K<=10000)组查询，查询有两个部分：C_i将是“P”或“Q”的命令。
如果C_i是‘P‘，则查询的第二部分将是一个整数A_i（1 <= A_i <= N！），它是行号。此时，你需要回答正确的牛线。
如果C_i是“Q”，则查询的第二部分将是N个不同的整数B_ij（1 <= B_ij <= N）。这将表示一条牛线，此时你需要输出正确的行号。
输入格式
* Line 1: Two space-separated integers: N and K
* Lines 2..2*K+1: Line 2*i and 2*i+1 will contain a single query.
Line 2*i will contain just one character: ‘Q‘ if the cows are lining up and asking Farmer John for their line number or ‘P‘ if Farmer John gives the cows a line number.
If the line 2*i is ‘Q‘, then line 2*i+1 will contain N space-separated integers B_ij which represent the cow line. If the line 2*i is ‘P‘, then line 2*i+1 will contain a single integer A_i which is the line number to solve for.
输出格式
* Lines 1..K: Line i will contain the answer to query i.
If line 2*i of the input was ‘Q‘, then this line will contain a single integer, which is the line number of the cow line in line 2*i+1.
If line 2*i of the input was ‘P‘, then this line will contain N space separated integers giving the cow line of the number in line 2*i+1.
输入输出样例
输入 #1复制
5 2 P 3 Q 1 2 5 3 4
输出 #1复制
1 2 4 3 5 5
说明/提示
感谢@prcups 提供翻译
这题题目中给了一个比较重要的信息
"很明显，问题的答案与奶牛进入谷仓的顺序无关。"
这句话告诉了我们这题的一个思路：将所有数据一起处理
我们可以先假设所有牛棚可以放无数个奶牛
然后一个一个往后推
这样得出的答案就是直接做的答案（当然牛棚和牛具体对应的序号是不知道的）
要注意可能爆int，以及第一遍扫过以后可能有的到了头一个牛棚，要再来一遍
#include<cstdio>
using namespace std;
long long int n,k,i,j,ans[3000005],x,y,a,b;
inline long long int read(){
long long int s=0,w=1;
char ch=getchar();
while(ch<‘0‘||ch>‘9‘){
if(ch==‘-‘){
w=-1;
}
ch=getchar();
}
while(ch>=‘0‘&&ch<=‘9‘){
s=s*10+ch-‘0‘;
ch=getchar();
}
return s*w;
}
int main(){
n=read();
k=read();
while(k--){
x=read();
y=read();
a=read();
b=read();
for(int i=1;i<=y;i++){
ans[(a*i+b)%n]+=x;
}
}
for(int i=0;i<n;i++){
if(ans[i]>0){
ans[(i+1)%n]+=ans[i]-1;
ans[i]=1;
}
}
while(ans[0]>1)
for(int i=0;i<n;i++){
if(ans[i]>0){
ans[(i+1)%n]+=ans[i]-1;
ans[i]=1;
}
}
for(int i=0;i<n;i++){
if(ans[i]==0){
printf("%lld\n",i);
return 0;
}
}
return 0;
}
原文地址：https://www.cnblogs.com/hrj1/p/11223359.html