AtCoder Beginner Contest 179 E - Sequence Sum (模拟)

• 题意:\(f(x,m)\)表示\(x\ mod\ m\),\(A_{1}=1\),而\(A_{n+1}=f(A^{2}_{n},M)\),求\(\sum^{n}_{i=1}A_{i}\).

• 题解:多算几个,会有一个循环节,直接模拟就好了.

• 代码:

  ll n,x,m;
  ll mp[N];
  vector<ll> a;
  
  int main() {
      //ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  	scanf("%lld %lld %lld",&n,&x,&m);
  	ll pos=n;
  	ll tmp=0;
  	for(ll i=1;i<=n;++i){
  		if(i>1) x=x%m*x%m;
  		tmp+=x;
  		if(!mp[x]){
  			mp[x]=i;
  			a.pb(x);
  		}
  		else{
  		   pos=i;
  		   break;
  		}
  	}
  	if(mp[x]==n){
  		printf("%lld\n",tmp);
  		return 0;
  	}
  	ll res=0;
  
  	for(ll i=0;i<mp[x]-1;++i){
  		res+=a[i];
  	}
  	ll cnt=0;
  	for(ll i=mp[x]-1;i<pos-1;++i){
  		cnt+=a[i];
  	}
  	ll len=pos-mp[x];
  	ll times=(n-mp[x]+1)/len;
  	ll rest=(n-mp[x]+1)%len;
  	res+=cnt*times;
  
  	for(ll i=mp[x]-1;i<mp[x]-1+rest;++i){
  		res+=a[i];
  	}
  
  	printf("%lld\n",res);
  
      return 0;
  }