ZOJ Bizarre Routine 分析上下界(2013 ACM/ICPC Asia Regional Changsha Online B)

Bizarre Routine

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Time Limit: 2 Seconds      Memory Limit: 32768 KB      Special Judge

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We've got a bizarre routine here:

bool less_than(x, y) {
	T++;
	return x < y;
}
void work(array[], l, r) {
	if (l >= r) return;
	swap(array[(l * A + r * B) / (A + B)], array[r]);
	int index = l;
	for (i = l; i < r; i++)
		if (less_than(array[i], array[r]))
			swap(array[index++], array[i]);
	swap(array[r], array[index]);
	work(array, l, index - 1);
	work(array, index + 1, r);
}
void main() {
	T = 0;
	Input(n, expect, A, B, array[]);
	work(array, 0, n - 1);
	if (T == expect)
		Output("YES");
	else
		Output("NO");
}

Now the question is: For given n, expect, A and B, how to arrange the input array to make the routine output "YES"?

To simplify the problem, we assume that the input array must be a permutation of 1 to n, and the division in the routine is integer division.

Input

There will be no more than 100 test cases, each case contains 4 integers n (1 ≤ n ≤ 10000), expect (1 ≤ expect ≤ 100000000), A (1 ≤ A ≤ 10) and B (1 ≤ B ≤ 10) in one line.

Output

If there exists an arrangement to make the routine output "YES", please print these corresponding n integers in one line, and seperate them by single spaces. Please be aware that these n integers must be a permutation of 1 to n and the answer might not be unique.

If there isn't any arrangement to make it output "YES", please print "NOWAY" in a line.

Sample Input

6 10 2 1
12 15 1 1

Sample Output

4 5 1 2 3 6
NOWAY

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 11015
#define mod 1000000007
#define INF 0x7fffffff
#define ll long long
//#define ll __int64
int p[maxn];
int mi[maxn];
int n,m,a,b;
void back(int l,int m,int r,int id){
    p[m]=m;swap(p[m],p[r]);swap(p[r],p[id]);
}
void work(int l,int r,int c){
    if(l>r)return;
    if(l==r){p[l]=l;return;}
    int l_l=l,l_r=((l+r)>>1)+1,m;
    c-=r-l;
    while(l_l<l_r){
        m=(l_l+l_r)>>1;
        if(mi[m-l]+mi[r-m]>c)l_l=m+1;
        else l_r=m;
    }
    m=l_l;
    work(l,m-1,mi[m-l]);
    work(m+1,r,c-mi[m-l]);
    back(l,m,r,(l*a+r*b)/(a+b));
}
int main(){
    for(int i=2;i<=maxn;i++){
        int x=i/2,y=i-x-1;
        mi[i]=mi[x]+mi[y]+i-1;
    }
    while(~scanf("%d%d%d%d",&n,&m,&a,&b)){
        if(m<mi[n]||m>(n-1)*(n)/2){printf("NOWAY\n");continue;}
        work(0,n-1,m);
        for(int i=0;i<n;i++)printf(i==n-1?"%d\n":"%d ",p[i]+1);
    }
    return 0;
}