Codeforces Round #547 (Div. 3) C. Polycarp Restores Permutation (数学)

• 题意:有一长度为\(n\)的序列\(p\),现在给你\(q_i=p_{i+1}-q_i \ (1\le i\le n)\),问你是否能还原出原序列,如果能救输出原序列,否则输出\(-1\).

• 题解:由:\(q_i=p_{i+1}-p_i\),我们对其求前缀和可得:\(s_i=p_{i+1}-p_1\),然后再求出:\(\sum^{n-1}_{i=1}s_i=\sum^{n}_{i=2}p_i-(n-1)*p_1\),\(\sum^{n}_{i=2}p_i=\sum^{n}_{i=1}i-p_1=\frac{(n)*(n+1)}{2}-p_1\),所以:\(\sum^{n-1}_{i=1}s_i= \frac{(n)*(n+1)}{2}-n*p_1\),直接推出\(p_1\),然后线性求出所有\(p_i\)再判断一下就好了.

• 代码:

  ll n;
  ll q[N],p[N];
  ll c[N];
  map<ll,int> mp;
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      cin>>n;
      rep(i,1,n-1) cin>>q[i];
  
      rep(i,1,n-1) c[i]=c[i-1]+q[i];
  
      ll cnt=0;
      rep(i,1,n-1) cnt+=c[i];
      ll sum=n*(n+1)/2;
      sum-=cnt;
      if(sum%n!=0){
      	cout<<-1<<'\n';
      	return 0;
      }
      p[1]=sum/n;
  
      rep(i,1,n-1){
      	p[i+1]=p[i]+q[i];
      }
      bool flag=true;
      rep(i,1,n){
      	if(mp[p[i]]) {flag=false;break;}
      	if(p[i]<=0 || p[i]>n) {flag=false;break;}
      	mp[p[i]]++;
      }
      if(!flag) cout<<-1<<'\n';
      else{
      	rep(i,1,n) cout<<p[i]<<' ';
      }
  
      return 0;
  }