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做法:
- si = (x+i)%p 由这个式子,可以推出x = p-1,p = max(ai)+1;
- 构造的话,暴力模拟一下,发现一定从0开始。我们需要做的就是贪心的往相邻两个数之间填充数即可。
AC代码:
#include<bits/stdc++.h>
#define IO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x) push_back(x)
#define sz(x) (int)(x).size()
#define sc(x) scanf("%d",&x)
#define pr(x) printf("%d\n",x)
#define abs(x) ((x)<0 ? -(x) : x)
#define all(x) x.begin(),x.end()
#define mk(x,y) make_pair(x,y)
#define debug printf("!!!!!!\n")
#define fin freopen("in.txt","r",stdin)
#define fout freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxm = 1e8+5;
const int maxn = 2e5+5;
const int INF = 0x3f3f3f3f;
ll a[maxn];
int main()
{
#ifdef LOCAL_FILE
fin;
#endif // LOCAL_FILE
//IO;
int n;
ll mx = -1;
sc(n);
for(int i=1;i<=n;i++)
{
sc(a[i]);
mx = max(mx,a[i]);
}
ll p = mx+1;
ll ans = n;
for(int i=2;i<=n;i++)
{
ll tmp = a[i]-a[i-1]-1;
if(tmp<0) ans+=tmp+p;
else ans+=tmp;
}
printf("%lld\n",ans);
return 0;
}