标签：== name sub ati i++ rest || contest its
给n-1个数，
qi=pi+1−pi
p为1-n的排列序列
q为给的序列。
根据q求出p， 求不出时为-1。
另p数组首个为0.求出p数组，再得到p数组中最小的值，将p中所有数减掉最小值再加1得到原p。
同时记录原p中每个数出现的次数。超过一次，说明不能还原 。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2e5 + 10;
LL a[MAXN];
map<LL, int> b;
int main()
{
int n, v;
cin >> n;
a[0] = 0;
LL sum = 0, mmin = 0;
for (int i = 2;i <= n;i++)
{
cin >> v;
a[i] = a[i - 1] + v;
sum += a[i];
mmin = min(mmin, a[i]);
}
for (int i = 1;i <= n;i++)
{
a[i] = a[i] - mmin + 1;
if (b[a[i]] == 1 || a[i] > n)
{
cout << -1 << endl;
return 0;
}
b[a[i]] = 1;
}
for (int i = 1;i <= n;i++)
cout << a[i] << ‘ ‘ ;
cout << endl;
return 0;
}
C. Polycarp Restores Permutation
标签：== name sub ati i++ rest || contest its
原文地址：https://www.cnblogs.com/YDDDD/p/10570961.html
