Lost Cows

* Line 1: A single integer, N
* Lines 2..N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on.

C++(g++ 7.5.0) 解法, 执行用时: 103ms, 内存消耗: 388K, 提交时间: 2023-04-09 11:59:11

#include<stdio.h>
int main() {
	int a[8000];
	int n;
	scanf("%d", &n);
	a[0] = 1;
	for (int i = 1; i < n; i++)
		scanf("%d", &a[i]);
	for (int i = 1; i < n; i++) {
		a[i] = a[i] + 1;
		for (int j = 0; j < i; j++)
			if (a[j] >= a[i])
				a[j] = a[j] + 1;
	}
	for (int i = 0; i < n; i++)
		printf("%d\n", a[i]);
	return 0;
}

pypy3 解法, 执行用时: 87ms, 内存消耗: 23612K, 提交时间: 2023-04-09 10:43:02

from sys import stdin, stdout
input = lambda: stdin.readline().strip()
print = lambda x: stdout.write(str(x) + '\n')


n = int(input())
lis = [1]
for i in range(2, n + 1):
    t = int(input())
    lis.insert(t, i)
dit = {}
for i in range(n):
    dit[lis[i]] = i + 1
for i in range(1, n + 1):
    print(dit[i])

C++11(clang++ 3.9) 解法, 执行用时: 41ms, 内存消耗: 480K, 提交时间: 2019-11-24 21:30:10

#include<stdio.h>
int a[8005],vis[8005],b[8005];
int main(){
	int n;
	scanf("%d",&n);
	a[1]=0;
	for(int i=2;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=n;i>=1;i--){
		int t=a[i]+1;
		for(int j=1;j<=t;j++)
			if(vis[j])
				t++;
		b[i]=t;
		vis[t]=1;
	}
	for(int i=1;i<=n;i++)
		printf("%d\n",b[i]);
}

Python3 解法, 执行用时: 324ms, 内存消耗: 5256K, 提交时间: 2023-04-09 10:39:56

a=int(input())
b=[]
for i in range(a-1):
    b.append(int(input()))
b=b[::-1]
c=list(range(1,a+1))
d=[]
for i in b:
    d.append(c[i])
    c.remove(c[i])
d.append(c[0])
d = d[::-1]
for i in range(a):
    print(d[i])

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