ztr loves math
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 766 Accepted Submission(s): 296
Problem Description ztr loves research Math.One day,He thought about the "Lower Edition" of triangle equation set.Such as n=x2−y2 .
He wanted to know that ,for a given number n,is there a positive integer solutions?
Input There are T test cases.
The first line of input contains an positive integer T(T<=106) indicating the number of test cases.
For each test case:each line contains a positive integer , n<=1018 .
Output If there be a positive integer solutions,print True ,else print False
Sample Input
4
6
25
81
105
Sample Output
False
True
True
True
Hint
For the fourth case,$105 = 13^{2}-8^{2}$
問題描述
ztr喜歡研究數學,一天,他在思考直角三角形方程組的Lower版,即n=x^{2}-y^{2}n=x2−y2,他想知道,對于給出的n,是否會有正整數解。
輸入描述 有T組資料,第一行為一個正整數T(T<=10^{6})T(T<=106),每一行一個正整數n,n <=10^{18}n<=1018
輸出描述 如果有正整數解,輸出TrueTrue,否則輸出FalseFalse
輸入樣例 4
6
25
81
105
輸出樣例 False
True
True
True
Hint 對于第四個樣例,有一組解13^{2}-8^{2}=105132−82=105
//首先可以模拟一下前面的幾個數: 前九個數的平方為: 1 4 9 16 25 36 49 64 81 相鄰兩個數的內插補點為: 3 5 7 9 11 13 15 17 由上面的數可以看出如果是相鄰兩個數的話,n的值為奇數,但也可能為偶數(比較特殊),例如: 9-1=8 16-4=12等... 除了這兩種情況沒有其他的情況了,是以根據這兩個發現可以找出規律: if((n&1ll&&n!=1)||(n%4==0&&n!=4))
printf("True\n");
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define ll long long
using namespace std;
int main()
{
int t,i,j,k;
ll n;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&n);
if((n&1ll&&n!=1)||(n%4==0&&n!=4))
printf("True\n");
else
printf("False\n");
}
return 0;
}