给逆序对数求原数组

给正序对数求原数组

假设 正序对数是 $01212$
解决:

1. 先管正序对数小的，从小往大放
2. 正序对数相同时，从右往左放

$00000$
$10000$
$13020$
$13524$

例题
这题需要根据残缺的数组，补齐原数组

#include<bits/stdc++.h>
#include<cstring>
#include<string>
using namespace std;
typedef long long ll;
#define SIZE (1000000+10)
int a[SIZE];
int b[SIZE];
int flag[SIZE];
vector<int>pot[SIZE];
int main() {
    int i,j,n,k,p,x;
    scanf("%d%d",&n,&k);
    for(i=0;i<k;i++){
        scanf("%d%d",&p,&x);
        b[p]=x;
    }
    int maxn=0;
    for(i=1;i<=n;i++){
        if(b[i]==0)b[i]=b[i-1]+1;
        else {
            if(b[i]>maxn+1){
                cout<<-1<<endl;
                return 0;
            }
        }
        maxn=max(maxn,b[i]);
        pot[b[i]].push_back(i);     
    }
    int ans=1;
    for(i=1;i<=n;i++){
        for(j=pot[i].size()-1;j>=0;j--){
            a[pot[i][j]]=ans++;
        }
    }
    //for(i=1;i<=n;i++)printf("%d ",b[i]);cout<<endl;
    for(i=1;i<=n;i++)printf("%d ",a[i]);cout<<endl;
   
    return 0;
}

给逆序对数求原数组（原数各不相同

排序
本题关键是得到原数组

1. 放入 $0012...x-1xx+1...n-2$
2. 多余的数从第二个位置开始加 ，可以得到逆序对数$012...xxx+1...n-2$
3. 然后再用当前位置扣去数字，可以得到正序对数$0011...10...0$
4. 先放0 ，再放1，从右往左放

#include<bits/stdc++.h>
using namespace std;
const int Size = 1e5 + 10;
#define ll long long
#define debug(x) {cout<<#x<<" = "<<x<<" ";}
#define debug1(a) {debug(a);cout<<endl;}
#define debug2(a,b) {debug(a);debug1(b);}
#define debug3(a,b,c) {debug(a);debug2(b,c);}
#define debugvec(x,n) {cout<<#x<<": ";for(int asd=0;asd<n;asd++)cout<<x[asd]<<" ";cout<<endl;}
#define debugvector(x) {cout<<#x<<": ";for(int asd=0;asd<x.size();asd++)cout<<x[asd]<<" ";cout<<endl;}
#define debugvectorstring(x) {cout<<#x<<": ";for (auto& s : x) printf("%s", s.c_str());;cout<<endl;}
int n, k = 100;
string s, ans;
vector<int>v;
void ini() {
    int i, t;
    n = 1;
    while (n * (n - 1) / 2 < k) n++;
    v.push_back(0);
    for (i = 0; i < n - 1; i++)v.push_back(i);
    t = k - (n - 2) * (n - 1) / 2;
    //debug3(k, (n - 2) * (n - 1) / 2, t);
    for (i = n - 1; t; t--, i--)v[i]++;
    for (i = 0; i < n; i++)v[i] = i - v[i];
    //debugvec(v, n);
}
void sovle() {
    int i;
    char ch = 'a';
    s.resize(n);
    for (i = n - 1; i >= 0; i--)if (!v[i]) s[i] = ch++;
    for (i = n - 1; i >= 0; i--)if (v[i]) s[i] = ch++;
    cout << s << endl;
}
int main() {
    //freopen("C:\\Users\\31531\\Desktop\\input.txt", "r", stdin);
    ini();
    sovle();
    return 0;
}

给逆序对数求原数组（原数可相同

字串排序

当$k\le \frac{26(26-1)}{2}=325时$

1. 放入 $0012...x-1xx+1...n-2$
2. 从倒二位开始$v[i]++$，每次往前挪两格$i-=2$
3. 从正二位或者正三位开始$v[i]++$，每次往后挪两格$i+=2$
4. 已经是递增的顺序了，直接从右往左放

#include<bits/stdc++.h>
using namespace std;
const int Size = 3e3 + 10;
#define ll long long
#define debug(x) {cout<<#x<<" = "<<x<<" ";}
#define debug1(a) {debug(a);cout<<endl;}
#define debug2(a,b) {debug(a);debug1(b);}
#define debug3(a,b,c) {debug(a);debug2(b,c);}
#define debugvec(x,n) {cout<<#x<<": ";for(int asd=0;asd<n;asd++)cout<<x[asd]<<" ";cout<<endl;}
#define debugvector(x) {cout<<#x<<": ";for(int asd=0;asd<x.size();asd++)cout<<x[asd]<<" ";cout<<endl;}
#define debugvectorstring(x) {cout<<#x<<": ";for (auto& s : x) printf("%s", s.c_str());;cout<<endl;}
string s = "";
int n,k=100;
vector<int>v;
void debugs(string s) {
	for (auto ch : s) {
		printf("%3c", ch);
	}
	cout << endl;
	int p[2] = { 0,0 };
	for (int i = 0; i < s.size(); i++) {
		int cnt = 0;
		for (int j = i - 1; j >= 0; j--) {
			if (s[j] > s[i]) {
				cnt++;
			}
		}
		p[1] += cnt;
		printf("%3d", cnt);
	}
	cout << endl;
	//debug2(p[0], p[1]);
}
void ini() {
	int i, t;
	n = 1;
	while (n * (n - 1) / 2 < k) n++;
	t = k - (n - 1) * (n - 2) / 2;
	v.push_back(0);
	for (i = 0; i < n - 1; i++)v.push_back(i);
	//debugvec(v, n);
	for (i = n - 2; i > 0 && t; i-=2, t--)v[i]++;
	//debugvec(v, n);
	for (i = 1 + (n % 2); t; t--, i += 2)v[i]++;
	//debugvec(v, n);
}
void solve() {
	s.resize(n);
	char ch = 'a';
	s[n - 1] = ch;
	for (int i = n - 2; i >= 0; i--) {
		if (v[i] == v[i + 1])s[i] = ch;
		else s[i] = ++ch;
	}
}
int flag = 26 * 25 / 2;
int main() {
	//freopen("C:\\Users\\31531\\Desktop\\input.txt", "r", stdin);
	//debugs("bbaa");
	//debugs("jihgfeeddccbbaa");
	cin >> k;
	if (k > flag)while (1);
	ini();
	solve();
	cout << s;
	return 0;
}

当$k>\frac{26(26-1)}{2}时$待续

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