HDU 5312 Sequence

Sequence

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 509    Accepted Submission(s): 126

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

10123456782210

 

Sample Output

1234561244

 

Source

BestCoder 1st Anniversary ($)

 

问题描述

Soda习得了一个数列, 数列的第n（n>=1）项是3n(n-1)+1. 现在他想知道对于一个给定的整数m, 是否可以表示成若干项上述数列的和. 如果可以, 那么需要的最小项数是多少?例如, 22可以表示为7+7+7+1, 也可以表示为19+1+1+1.

输入描述

输入有多组数据. 第一行有一个整数T(1<=T<=10000), 表示测试数据组数. 然后对于每组数据:一行包含1个整数 m(1<= m <=10^9).

输出描述

对于每组数据输出最小花费.

输入样例

10123456782210

输出样例

1234561244

解题思路：3n(n-1)+1=3*2*(n(n-1)/2)+1=6*(n(n-1)/2)+1，任意一个自然数最多只需要3个三角形数即可表示。而n(n-1)/2（n>=1）正是三角形数的通项公式，因此n(n-1)/2（n>=1）一定是整数。假设m是k个题目中的数列的数的和，则m=6*(k个三角形数的和)+k。题目要求是找到最小的k（k>=1），那么只需找到满足条件（m-k）%6==0的最小k即可，但是对于k=1和k=2的特殊情况要特判处理。

代码如下：

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<ctime>#include<iostream>#include<algorithm>#include<string>#include<vector>#include<deque>#include<list>#include<set>#include<map>#include<stack>#include<queue>#include<numeric>#include<iomanip>#include<bitset>#include<sstream>#include<fstream>#include<limits.h>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-4)#define inf (1<<28)#define sqr(x) (x) * (x)#define mod 1e9+7using namespace std;typedef long long ll;typedef unsigned long long ULL;int n,f[100005];void Init(){    for(int i=1;;i++)    {        f[i]=3*i*(i-1)+1;        if(f[i]>1e9)        {            n=i;            break;        }    }}//一个数/*//调用函数二分查找int find_1(int m){  int k=lower_bound(f+1,f+1+n,m)-f;  return f[k]==m;}*//*//遍历查找int find_1(int m){    for(int i=1;i<=n;i++)        if(f[i]==m)        return 1;    return 0;}*///二分查找int find_1(int m){    int l=1,r=n;    int mid=(l+r)/2;    while(l<=r)    {        if(f[mid]==m)            return 1;        else if(f[mid]<m)            l=mid+1;        else            r=mid-1;        mid=(l+r)/2;    }    return 0;}//两个数/*int find_2(int m){  int l=1,r=n;  while(l<=r)  {     if(f[l]+f[r]<m)         return 0;     while(l<=r)     {         if(f[l]+f[r]<m)         {             r++;             break;        }        else if(f[l]+f[r]==m)            return 1;        else            r--;     }     l++;  }  return 0;}*///二分查找int find_2(int m){    int i,j;    for(i=1,j=n;i<=n&&f[i]<m;i++)    {        while(j>0&&f[i]+f[j]>m)            j--;        if(j>0&&f[i]+f[j]==m)            return 1;    }    return 0;}int main(){    Init();    int i,t,m;    scanf("%d",&t);    while(t--)    {        scanf("%d",&m);        if(find_1(m))            printf("1\n");        else        {            int flag=0;            if((m-2)%6==0)            {                if(find_2(m))                    flag=1;            }            if(flag)            printf("2\n");            else            {                 for(i=3;i<10;i++)                 {                      if((m-i)%6==0)                      {                           printf("%d\n",i);                           break;                      }                 }            }        }    }    return 0;}