hdu 5312 Sequence（数学推导+线性探查（两数相加版））

Problem Description

Today, Soda has learned a sequence whose n-th (n≥1) item is 3n(n−1)+1. Now he wants to know if an integer m can be represented as the sum of some items of that sequence. If possible, what are the minimum items needed?

For example, 22=19+1+1+1=7+7+7+1.

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤104), indicating the number of test cases. For each test case:

There's a line containing an integer m (1≤m≤109).

 

Output

For each test case, output −1 if m cannot be represented as the sum of some items of that sequence, otherwise output the minimum items needed.

 

Sample Input

10
1
2
3
4
5
6
7
8
22
10

 

Sample Output

1
2
3
4
5
6
1
2
4
4

 

Source

BestCoder 1st Anniversary ($)

 

 对于这种题，首先一开始就要对给出的公式进行研究，从这里入手是正道。

分析公式 3n(n−1)+1 ，若给出一个数n，假设要k个3n(n−1)+1加起来得到n，即 3n(n-1)k+k=n,注意到3n(n-1)k是6的倍数，那么求的是满足（n-k）%6==0的最小的k。还有就是要注意到1、2要特判，即k要从3开始枚举

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#include<stdlib.h>
#include<map>
using namespace std;
#define N 20000
int num[N];
int n;
void init(){
    for(int i=1;i<N;i++){
        num[i]=3*i*(i-1)+1;
    }
}
bool check1(){
    for(int i=1;i<N;i++){
        if(num[i]==n)
          return true;
    }
    return false;
}
bool check2(){
    int j=N-1;
    for(int i=1;i<N;i++){
        while(num[i]+num[j]>n && j>0) j--;
        if(num[i]+num[j]==n && j>0){
            return true;
        }
    }
    return false;
}
int main()
{
    init();
    int t;
    scanf("%d",&t);
    while(t--){
        
        scanf("%d",&n);
        if(check1()){
            printf("1
");
        }
        else if(check2()){
            printf("2
");
        }
        else{
            for(int i=3;i<10;i++){
                if((n-i)%6==0){
                    printf("%d
",i);
                    break;
                }
            }
        }
    }
    return 0;
}