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≥) item is 3n(n−)+. 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, =+++=+++.

Input

There are multiple test cases. The first line of input contains an integer T (≤T≤), 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 − if m cannot be represented as the sum of some items of that sequence, otherwise output the minimum items needed.

 

Sample Input

 

Sample Output

 

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=;i<N;i++){
         num[i]=*i*(i-)+;
     }
 }
 bool check1(){
     for(int i=;i<N;i++){
         if(num[i]==n)
           return true;
     }
     return false;
 }
 bool check2(){
     int j=N-;
     for(int i=;i<N;i++){
         while(num[i]+num[j]>n && j>) j--;
         if(num[i]+num[j]==n && j>){
             return true;
         }
     }
     return false;
 }
 int main()
 {
     init();
     int t;
     scanf("%d",&t);
     while(t--){

         scanf("%d",&n);
         if(check1()){
             printf("1\n");
         }
         else if(check2()){
             printf("2\n");
         }
         else{
             for(int i=;i<;i++){
                 if((n-i)%==){
                     printf("%d\n",i);
                     break;
                 }
             }
         }
     }
     return ;
 }