Heron and His Triangle（HDU 6222 找规律+大数）

Heron and His Triangle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 751    Accepted Submission(s): 364

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

41234

 

Sample Output

4444

 

Source

2017ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）


//题意：
给定一个n, 求t-1，t，t+1作为3条边构成的三角形面积为正整数，t>=n，求最小t 。

//思路：
三角形面积 S=(1/2)*a*b*sin<a,b> ;
cos<a,b> = (a^2+b^2-c^2)/(2*a*b) ;
sin<a,b>^2+cos<a,b>^2=1 ;
a=a; b=a+1; c=a+2;

=>S = (a+1)*sqrt(3*(a-1)*(a+3))/4 ; S要为正整数 ;

=>写个小程序枚举满足条件的t，得：t = 4, 14, 52, 194, 724 ...

=>规律： t[i]=t[i-1]*4-t[i-2] ;

按照上述规律，大数打表即可：由于t大致是按照4的指数上升的，所以10^32内，t大致是150个左右。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <vector>#include <algorithm>using namespace std;char num[500][40];//大数乘法void mul(char a[], int x,char (&res)[40]){    int len=strlen(a);    int flag=0;    char ans[40];    for(int i=len-1;i>=0;i--)    {        int val=(a[i]-'0')*4+flag;        ans[i]=val%10+'0';        flag=val/10;    }    ans[len]='\0';    res[0]='\0';    if(flag)    {        res[0]=flag+'0';        res[1]='\0';        strcat(res,ans);    }    else    {        strcpy(res,ans);    }}//大数减法void sub(char (&a)[40], char b[]){    char ans[40];    char tmp[40];    int lena=strlen(a);    int lenb=strlen(b);    for(int i=0;i<lena-lenb;i++)    {        ans[i]='0';    }    ans[lena-lenb]='\0';    strcat(ans,b);    strcpy(tmp,b);    strcpy(b,ans);    int len=max(lena,lenb);    int flag=0;    memset(ans,0,sizeof(ans));    for(int i=len-1;i>=0;i--)    {        int val=a[i]-b[i]-flag;        if(val<0)        {            ans[i]=val+10+'0';            flag=1;        }        else        {            ans[i]=val+'0';            flag=0;        }    }    ans[len]='\0';    strcpy(a,ans);    strcpy(b,tmp);}int judge(char a[]){    int len=strlen(a);    if(len>=32)        return 1;    else        return 0;}//比较两个数大小int compare(char a[], char b[]){    int lena=strlen(a);    int lenb=strlen(b);    if(lena>lenb)        return 1;    else if(lena<lenb)        return 0;    for(int i=0;i<lena;i++)    {        if(a[i]>b[i])            return 1;        else if(a[i]<b[i])            return 0;    }    return 2;}int main(){    int T;    char n[40];    //打表    strcpy(num[0],"4");    strcpy(num[1],"14");    for(int i=2;;i++)    {        mul(num[i-1],4,num[i]);        sub(num[i],num[i-2]);        if(judge(num[i]))            break;    }    scanf("%d",&T);    while(T--)    {        scanf("%s",n);        if(strcmp(n,"1")==0||strcmp(n,"2")==0||strcmp(n,"3")==0||strcmp(n,"4")==0)        {            printf("4\n");            continue;        }        int flag=0;        int cnt;        for(int i=0;;i++)        {            int x=compare(n,num[i]);            if(flag==1&&x!=1)            {                cnt=i;                break;            }            if(x==1)                flag=1;        }        printf("%s\n",num[cnt]);    }    return 0;}