Codeforces Round #702 (Div. 3)

B. Balanced Remainders

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[30005];
ll b[100005];
int main(){
    ll t;
    cin>>t;
    while(t--){
        ll n;
        cin>>n;
        memset(b,0,sizeof(b));
        for(ll i=1;i<=n;i++){
            cin>>a[i]; 
            if(a[i]%3==0) b[0]++;
            else if(a[i]%3==1) b[1]++;
            else b[2]++;
        }
        int num=0;
        for(int i=0;i+1<=100;i++){
            if(b[i]>b[i+1]){
                int pre=b[i];
                int x=(b[i]+b[i+1]);
                if(x%2==0) b[i+1]=x/2;
                else b[i+1]=x/2+1;
                b[i]=x-b[i+1];
                num+=abs(pre-b[i]);
            }
            b[i+3]=b[i];
        }
        cout<<num<<endl;
    }
    return 0;
}

C. Sum of Cubes
用map从1到1e4打表，然后for循环判断即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
map<ll,ll> mp;
int main(){
    //把1-10^12里面的立方求出来
    for(ll i=1;i<=10000;i++) mp[i*i*i]=1;
    ll t;
    cin>>t;
    while(t--){
        ll x;
        cin>>x;
        bool flag=false;
        for(ll i=1;;i++){
            if(x-i*i*i<=0) break;
            if(mp[x-i*i*i]==1){
                flag=true;
                cout<<"YES"<<endl;
                break;
            }
        }
        if(!flag) cout<<"NO"<<endl;
    }
    return 0;
}