K. Perpetuum Mobile
Time Limit: 2 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100187/problem/K
Description
The world famous scientist Innokentiy almost finished the creation of perpetuum mobile. Its main part is the energy generator which allows the other mobile's parts work. The generator consists of two long parallel plates with n lasers on one of them and n receivers on another. The generator is considered to be working if each laser emits the beam to some receiver such that exactly one beam is emitted to each receiver.
It is obvious that some beams emitted by distinct lasers can intersect. If two beams intersect each other, one joule of energy is released per second because of the interaction of the beams. So the more beams intersect, the more energy is released. Innokentiy noticed that if the energy generator releases exactly k joules per second, the perpetuum mobile will work up to 10 times longer. The scientist can direct any laser at any receiver, but he hasn't thought of such a construction that will give exactly the required amount of energy yet. You should help the scientist to tune up the generator.
Input
The only line contains two integers n and k (1 ≤ n ≤ 200000, ) separated by space — the number of lasers in the energy generator and the power of the generator Innokentiy wants to reach.
Output
Output n integers separated by spaces. i-th number should be equal to the number of receiver which the i-th laser should be directed at. Both lasers and receivers are numbered from 1 to n. It is guaranteed that the solution exists. If there are several solutions, you can output any of them.
Sample Input
4 5
Sample Output
4 2 3 1
HINT
题意
给你n个数，让你构造存在k个逆序数的序列
题解：
查找方法 n-1..............下去
代码
1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <ctime> 5 #include <iostream> 6 #include <algorithm> 7 #include <set> 8 #include <vector> 9 #include <queue> 10 #include <map> 11 #include <stack> 12 #define MOD 1000000007 13 #define maxn 32001 14 using namespace std; 15 typedef __int64 ll; 16 inline ll read() 17 { 18 ll x=0,f=1;char ch=getchar(); 19 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 20 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 21 return x*f; 22 } 23 //******************************************************************* 24 ll a[200101]; 25 ll q[200101]; 26 int main() 27 { 28 ll n; 29 ll k; 30 n=read(),k=read(); 31 for(ll i=1; i<=n; i++) 32 a[i]=i; 33 if(k==0) 34 { 35 for(ll i=1; i<=n; i++) 36 printf("%I64d ",a[i]); 37 return 0; 38 } 39 ll j=1; 40 ll x=0; 41 ll kk=1; 42 ll nn=n; 43 while(k) 44 { 45 if(k>=n-j) 46 { 47 q[++x]=a[nn]; 48 nn--; 49 k=k-(n-j); 50 } 51 else 52 { 53 q[++x]=a[kk]; 54 kk++; 55 } 56 j++; 57 } 58 j--; 59 for(ll i=1; i<=x; i++) 60 { 61 printf("%I64d ",q[i]); 62 } 63 for(ll i=kk; i<nn+1; i++) 64 printf("%I64d ",a[i]); 65 return 0; 66 }