本文主要是介绍HDU4206-巧妙推理，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目：题目链接
题意：对于直角三角形三边:x^2-y^2=z^2;题目中给出z^2，要求出最小的一组整数x和y。
分析：
1：将等式变形(x-y)(x+y)=z^2;
2：令A=x-y,B=x+y,T=z^2;于是A*B=T; x=(B+A)/2,y=(B-A)/2;
3：A和B是 T 的两个约数,只要A和B同时为奇数或者同时为偶数(B-A是偶数)，就能保证x,y是整数;
4：找最小的一组，从sqrt(T)往下开始找T的约数，先找到的满足前面条件的就是最小的了。
巧妙的变幻推理：
#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int gcd(int n,int m)
{if(n<m) swap(n,m);return n%m==0?m:gcd(m,n%m);
}
int lcm(int n,int m)
{if(n<m) swap(n,m);return n/gcd(n,m)*m;
}
#define N 10000007
int prime[N];
struct node
{int x, y;
};
bool cmp(const node & a, const node & b)
{return a.x > b.x;
}
void getPrime();
void bash();
void wzf();
void SG();
int QuickMod(int a, int b, int n);
int t, n;
int main()
{scanf("%d", &t);while(t--){scanf("%d", &n);int tt =sqrt(1.0*n);int fp = 0;for(int i = tt; i >= 1; --i){if(n % i==0){int a = i;int b = n/a;if((a%2==1 && (b%2 == 1)) || (a%2==0 && b %2 == 0)){//cout << "a: " << a << " b: " << b << endl;int x = (b-a)/2;int y = (a+b)/2;fp = 1;printf("%d %d\n", x, y);break;}}}if(!fp)printf("IMPOSSIBLE\n");}return 0;
}int QuickMod(int a,int b,int n)
{int r = 1;while(b){if(b&1)r = (r*a)%n;a = (a*a)%n;b >>= 1;}return r;
}void getPrime()
{memset(prime, 0, sizeof(prime));prime[0] = 1;prime[1] = 1;for(int i = 2; i < N; ++i){if(prime[i] == 0){for(int j = i+i; j < N; j+=i)prime[j] = 1;}}
}void bash(int n, int m)
{if(n%(m+1) != 0)printf("1\n");elseprintf("0\n");
}void wzf(int n, int m)
{if(n > m)swap(n, m);int k = m-n;int a = (k * (1.0 + sqrt(5.0))/2.0);if(a == n)printf("0\n");elseprintf("1\n");
}int numsg[N];
void SG(int n)
{int sum = 0;for(int i=0; i < n; i++){scanf("%d",&numsg[i]);sum ^= numsg[i];}if(sum == 0)printf("No\n");else{printf("Yes\n");for(int i = 0; i < n; i++){int s = sum ^ numsg[i];if(s < numsg[i])printf("%d %d\n", numsg[i], s);//从有num[i]个石子的堆后剩余s个石子}}
}
这篇关于HDU4206-巧妙推理的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
