Face questions
【Problem Description】
Dad came with a small S JZ Square, we found a small S side of the square from left to right with a row of pillars of different heights, very spectacular. Dad told small S, where a total of N columns, each column height is an integer between 1 and N, and any two columns is not as high.
Small S just learned to count and compare the size of two integers, and she would like to take this opportunity to show off J. She find a piece of paper, the write sequence number N, where the i-th from the left represents the number of i left columns higher than its column. For example, if arrangement is 3,1,2 column, a note will be so small S 0,1,1.
Now a small S would like to test you, she put the paper to you, you can know how the column is arranged it?
[Input Format]
The first row contains an integer \ (N \) ( \ (. 1 \ Le N \ 30000 Le \) ), represents the number of columns.
The next \ (N \) lines, each line an integer. The first \ (i + 1 \) line represents the paper of \ (I \) number.
[Output format]
The first line is YES, or NOindicating whether the number of the small S.
If the first line YES, the following output \ (N \) lines, each line an integer. The first \ (i + 1 \) line from the left represents the section \ (I \) height of the pillars.
Sample input [1]
7
0
0
1
0
2
0
4 Sample output [1]
YES
1
5
2
6
4
7
3 Sample input [2]
3
0
2
0Sample output [2]
NOsolution
Not difficult to find, which is equivalent to seeking ranked \ (a_i \) number, and deleting it. We use a std::vectormaintenance unused height, then every time you can find this number and delete.
program
#define MAXN 30050
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int n;
int a[MAXN], r[MAXN];
vector<int> ch;
void fail() {
cout << "NO" << endl;
exit(0);
}
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
ch.push_back(i + 1);
}
for (int i = n - 1; i > -1; i--) {
int ind = ch.size() - a[i] - 1;
// cout << "ind " << ind << ", size " << ch.size() << ", no. " << a[i] << endl;
// for (int j = 0; j < ch.size(); j++) cout << ch[j] << " "; cout << endl;
if (ind < 0 || ind >= ch.size()) fail();
r[i] = ch[ind];
if (r[i] < 1) fail();
// bitadd(r[i], 1);
ch.erase(ch.begin() + ind); // error?
}
cout << "YES" << endl;
for (int i = 0; i < n ; i++) {
cout << r[i] << endl;
}
return 0;
}