E. Singers’ Tour

我们显然可以推式子
b[i] = a[i]+2a[i+1]+3a[i+1]….
b[i+1]=na[i]+a[i+1]+2a[i+2]….
这样
b[i+1]-b[i]=-n*a[i]+(a[i]+a[i+1]+….+a[n])
我们显然可以直接计算出整个(a[i]+a[i+1]+….+a[n])
最后注意判断不能为小数负数 sum也要

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6+10;
const int M = 998244353;
const int mod = 1e9+7;
#define int long long
int up(int a,int b){return a<0?a/b:(a+b-1)/b;}
#define endl '\n'
#define all(x) (x).begin(),(x).end()
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define _ 0
#define pi acos(-1)
#define INF 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);

void solve(){
    int n;cin>>n;
    vector<int>b(n+10),a(n+10);
    int sum=0;
    for(int i=0;i<n;i++)cin>>b[i],sum+=b[i];
    vector<int>v(n+10);
    if(n==1){
        YES
        cout<<b[0]<<endl;
        return;
    }
    int cnt=1,now=0;
    while(cnt!=n+1)now+=cnt++;
    if(sum%now){NO return;}
    sum/=now;
    for(int i=0;i<n;i++){
        v[i]=(b[(i+1)%n]-b[i]);
    }
    vector<int>ans;
    for(int i=0;i<n;i++) {
        ans.push_back((v[i] - sum) / (-n));
        if((v[i] - sum) % (-n)){NO return;}
        if(ans[i]<=0){NO return;}
    }
    YES
    cout<<ans.back()<<' ';
    for(int i=0;i<ans.size()-1;i++)cout<<ans[i]<<' ';
    cout<<endl;
}
signed main(){
    fast
    int t;t=1;cin>>t;
    while(t--) {
        solve();
    }
    return ~~(0^_^0);
}

原文地址：http://www.cnblogs.com/ycllz/p/16814739.html