HDU6222 Heron and His Triangle(大数运算+规律)

Problem Description

A triangle is a Heron’s triangle if it satisfies that the side lengths of it are consecutive integers t-1,t,t+1 and thatits area is an integer. Now, for given n you need to find a Heron’s triangle associated with the smallest t bigger
than or equal to n.

Input

The input contains multiple test cases. The first line of a multiple input is an integer T (1 ≤ T ≤ 30000) followedby T lines. Each line contains an integer N (1 ≤ N ≤ 10^30).

Output

For each test case, output the smallest t in a line. If the Heron’s triangle required does not exist, output -1.

Sample Input

4

1

2

3

4

Sample Output

4

4

4

4

思路

因为刚开始没思路，所以试着写了个小程序，找出了符合条件的几种情况

发现 当 t=4,14,52,194 的时候满足题目要求，然后试图寻找数据之间的关系，可以发现t[i]=4*t[i-1]-t[i-2] (t>=2)

接下来就是递推，大数运算

代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAX 1000
char ans[MAX][MAX]={"0","4","14","52","194"};
void Multiplication(int pos)
{
    int x[MAX],r[MAX],i,k;
    int l=strlen(ans[pos-1]);
    memset(r,0,sizeof(r));
    for(i=l-1,k=0;i>=0;i--,k++)
        x[k]=ans[pos-1][i]-'0';
    for(i=0;i<l;i++)
        r[i]=x[i]*4;
    for(i=0;i<l+5;i++)
    {
        if(r[i]>=10)
        {
            r[i+1]+=r[i]/10;
            r[i]%=10;
        }
    }
    i=l+5;
    while(!r[i])
    {
        i--;
    }
    ans[pos][i+1]='\0';
    int index=0;
    while(i>=0)
    {
       ans[pos][index]=r[i]+'0';
       i-=1; index+=1;
    }
}
void Substraction(int pos)
{
    int len_re=strlen(ans[pos]);
    int len_Minus=strlen(ans[pos-2]);
    int i,k,A[MAX],B[MAX];
    memset(A,0,sizeof(A));
    memset(B,0,sizeof(B));
    for(i=len_re-1,k=0;i>=0;i--,k++)
        A[k]=ans[pos][i]-'0';
    for(i=len_Minus-1,k=0;i>=0;i--,k++)
        B[k]=ans[pos-2][i]-'0';
    for(i=0;i<len_re;i++)
    {
        A[i]-=B[i];
        if(A[i]<0)
        {
            A[i]+=10;
            A[i+1]--;
        }
    }
    len_re+=2;
    while(!A[len_re])
    {
        len_re-=1;
    }
    ans[pos][len_re+1]='\0';
    int index=0;
    while(len_re>=0)
    {
        ans[pos][index]=A[len_re]+'0';
        len_re-=1; index+=1;
    }
}
void cal()
{
    for(int i=5;i<=300;i++)
    {
        Multiplication(i);
        Substraction(i);
    }
}
int compare(char *str,int pos)
{
    int l_str=strlen(str);
    int l=strlen(ans[pos]);
    if(l_str<l)
        return 1;
    else if(l_str>l)
        return 0;
    for(int i=0;i<l;i++)
    {
        if(str[i]>ans[pos][i])
            return 0;
        else if(str[i]<ans[pos][i])
            return 1;
    }
    return 1;
}
int main()
{
    int T;
    scanf("%d",&T);
    cal();
    while(T--)
    {
        char str[MAX];
        scanf("%s",str);
        for(int i=1;i<=300;i++)
        {
            if(compare(str,i))
            {
               printf("%s\n",ans[i]);
               break;
            }
        }
    }
    return 0;
}
//大数+数学规律