hdu 5312 Sequence

Sequence

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 878    Accepted Submission(s): 267

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 integerT(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

官方题解：

这个题看上去是一个贪心, 但是这个贪心显然是错的.事实上这道题目很简单, 先判断1个是否可以, 然后判断2个是否可以. 之后找到最小的$k(k>2)$, 使得$(m-k)mod6=0$即可.

证明如下:$3n(n-1)+1=6(n\ast (n-1)/2)+1$, 注意到$n\ast (n-1)/2$是三角形数, 任意一个自然数最多只需要3个三角形数即可表示. 枚举需要$k$个, 那么显然$m=6(k$个三角形数的和$)+k$, 由于$k\ge 3$, 只要$m-k$是6的倍数就一定是有解的.

事实上, 打个表应该也能发现规律.

尽然还用到三角形数。。看了之后枚举还是超时，，后来才懂原来是在枚举2的时候超时了，开始的做法是枚举每个数与n的差值，用二分找差值，如果存在就枚举成功，时间复杂度是nlogn,但是还是超时（T比较大）。后来想到原数列是有序的，所以从原数列中找两个数的和为n可以用线性的算法。设置两个指针 l 和r 一个指向最前面，一个指向最后面，l指针不断往后，r指针不断往前，由于是排序好的，两个指针都不需要回溯。所以比较快。

#include<map>#include<vector>#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<stack>#include<queue>#include<set>#define inf 0x3f3f3f3f#define mem(a,x) memset(a,x,sizeof(a))using namespace std;typedef long long ll;typedef pair<int,int> pii;inline ll in(){    ll res=0;char c;    while((c=getchar())<'0' || c>'9');    while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();    return res;}int a[20005];int main(){    mem(a,0);    for(int i=1;i<=20000;i++)    {        a[i]=3*i*(i-1)+1;    }    int T=in();    while(T--)    {        int n=in();        int pos=lower_bound(a+1,a+20000,n)-a;        if(a[pos]==n)          //枚举1        {            puts("1");            continue;        }        int l=1,r=pos-1;        bool ok=0;        while(l<=r)    //枚举2        {            while(a[l]+a[r]>n)r--;            if(a[l]+a[r]==n)            {                ok=1;                break;            }            ++l;        }        if(ok)        {            puts("2");            continue;        }        for(int i=3;i<=8;i++)//枚举3-8        {            if((n-i)%6==0)            {                printf("%d\n",i);                break;            }        }    }    return 0;}