题目描述
题目分析
首先想到的方法为dfs去寻找每一个数，但发现会有超时
cpp
#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, cnt, a[N];
void dfs(int dep, int sum, int start)
{
if(dep == 4)
{
if(sum == 0 && cnt == 0)
{
for(int i = 0; i < 4; i ++)
{
cout << a[i] << ' ';
}
cnt ++;
}
return;
}
for(int i = start; i <= sqrt(sum); i ++)
{
a[dep] = i;
dfs(dep + 1, sum - (i * i), i);
}
}
int main()
{
cin >> n;
dfs(0, n, 0);
return 0;
}使用二分
先将后两个数确定，将其后两个数的平方和以及分别对应的数字存入结构体中，再一一枚举前两个数，二分出可以匹配的后两个数，确定出答案
cpp
#include<bits/stdc++.h>
using namespace std;
const int N = 5e6 + 10;
int n, num;
struct node
{
int ss, c, d;
}sum[N * 2];
bool cmp(node x, node y)
{
if(x.ss != y.ss)
{
return x.ss < y.ss;
}
else
{
if(x.c != y.c)
{
return x.c < y.c;
}
else
{
return x.d < y.d;
}
}
}
int main()
{
cin >> n;
for(int c = 0; c * c <= n; c ++)
{
for(int d = c; c * c + d * d <= n; d ++)
{
sum[num ++] = {c * c + d * d, c, d};
}
}
sort(sum, sum + num, cmp);
for(int a = 0; a * a <= n; a ++)
{
for(int b = 0; a * a + b * b <= n; b ++)
{
int t = n - a * a - b * b;
int l = 0, r = num - 1;
while(l < r)
{
int mid = l + r >> 1;
if(sum[mid].ss >= t)r = mid;
else l = mid + 1;
}
if(sum[l].ss == t)
{
cout << a << ' ' << b << ' ' << sum[l].c << ' ' << sum[l]. d;
return 0;
}
}
}
return 0;
}