hdu5312

看官方题解可知：
此题元素组成的公式为3n*(n-1)+1,如果m是由其k个元素组成，那么m = 3*n*(n-1)k+k，而我们发现，3*n(n-1)肯定是一个6的倍数，而还要注意我们要特殊判断m是否能由1个或者2个元素组成，当元素数目大于等于3时，我们再用m = 3*n*(n-1)*k+k判断，化简之，就是(m-k)%6 == 0,找到最小的k即可。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define inf 0x3f3f3f3f#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define ll __int64const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;int n,m;#define N 20010#define Mod 1000000007#define maxn 1010char ch[110];ll a[N];void Init(){    for(ll i=1;i<N;i++){        a[i] = 3*i*(i-1)+1;    }}bool check1(){    for(int i=1;i<N;i++)        if(a[i] == m)            return true;    return false;}bool check2(){    for(int i=1,j=N-1;i<N&&a[i]<m;i++){        while(j>0 && a[i]+a[j]>m){            j--;        }        if(j>0 && a[i]+a[j] == m){            return true;        }    }    return false;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif   int t;   Init();   sf(t);   while(t--){       sf(m);       if(check1())           printf("1\n");       else if(check2()){           printf("2\n");       }else{           int ans;           for(int i=3;i<10;i++){               if((m-i)%6 == 0){                   ans = i;                   break;               }           }           printf("%d\n",ans);       }   }return 0;}