版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/a54665sdgf/article/details/81434771
题目：
Consider equations having the following form:
a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.
It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.
Determine how many solutions satisfy the given equation.
（懒得翻译了，直接抄原文）
这道题的解法非常多，光我见到的就至少得有五六种。就是数据比较刁钻，十组得有9组是全正或全负的，如果不加个判断，许多理论复杂度可行的方法都会TLE，囧~~~
我简单地介绍几种方法：
解法一(530ms)：三重循环暴力+整除判断
枚举x1,x2,x3，并用等式性质算出x4，并判断是否在范围内。（因为正负平方会重复，所以只取1-100内的值，将结果乘16即可）
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#define FRER() freopen("i.txt","r",stdin)
#define FREW() freopen("o.txt","w",stdout)
using namespace std;
typedef long long LL;
const int N=10000+100;
int vis[N];
int main()
{
//FRER();
memset(vis,0,sizeof vis);
for(int i=1; i<=100; ++i)
{
vis[i*i]=1;
}
int a,b,c,d;
while(cin>>a>>b>>c>>d)
{
if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
{
printf("0\n");
continue;
}
int cnt=0;
for(int x1=1; x1<=100; ++x1)
for(int x2=1; x2<=100; ++x2)
for(int x3=1; x3<=100; ++x3)
{
int p=-(a*x1*x1+b*x2*x2+c*x3*x3);
if(p%d==0&&p/d>0&&p/d<=10000&&vis[p/d])++cnt;
}
cout<<cnt*16<<endl;
}
return 0;
}解法二(764ms)：用sqrt直接判断是否为平方数，省去了vis数组的空间开销，其他同解法一
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#define FRER() freopen("i.txt","r",stdin)
#define FREW() freopen("o.txt","w",stdout)
using namespace std;
typedef long long LL;
bool ok(int x)
{
return abs(sqrt(x)-round(sqrt(x)))<1e-5;
}
int main()
{
//FRER();
int a,b,c,d;
while(cin>>a>>b>>c>>d)
{
if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
{
printf("0\n");
continue;
}
int cnt=0;
for(int x1=1; x1<=100; ++x1)
for(int x2=1; x2<=100; ++x2)
for(int x3=1; x3<=100; ++x3)
{
int p=-(a*x1*x1+b*x2*x2+c*x3*x3);
if(p%d==0&&p/d>0&&p/d<=10000)
{
if(ok(p/d))++cnt;
}
}
cout<<cnt*16<<endl;
}
return 0;
}解法三(280ms)：折半枚举，先枚举x1,x2的值，用map记录ax1^2+bx2^2出现的次数，再枚举x3,x4的值，将结果加上-(cx3^2+dx4^2)出现的次数。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#include<map>
#define FRER() freopen("i.txt","r",stdin)
#define FREW() freopen("o.txt","w",stdout)
using namespace std;
typedef long long LL;
const int N=2000000+100;
map<int,int> cnt;
int main()
{
//FRER();
int a,b,c,d;
while(scanf("%d%d%d%d",&a,&b,&c,&d)==4)
{
if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
{
printf("0\n");
continue;
}
cnt.clear();
int ans=0;
for(int x1=1; x1<=100; ++x1)
for(int x2=1; x2<=100; ++x2)
cnt[-(a*x1*x1+b*x2*x2)]++;
for(int x3=1; x3<=100; ++x3)
for(int x4=1; x4<=100; ++x4)
if(cnt.count(c*x3*x3+d*x4*x4))ans+=cnt[c*x3*x3+d*x4*x4];
printf("%d\n",ans*16);
}
return 0;
}解法四(31ms)：用cnt数组来保存枚举过的值，并用另一个Index数组来记录出现过的值，清零的时候只清零所有Index的cnt值即可，其余同解法三。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#define FRER() freopen("i.txt","r",stdin)
#define FREW() freopen("o.txt","w",stdout)
using namespace std;
typedef long long LL;
const int N=2000000+100;
int cnt[N],Index[200000],k=0;
int main()
{
//FRER();
int a,b,c,d;
while(scanf("%d%d%d%d",&a,&b,&c,&d)==4)
{
if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
{
printf("0\n");
continue;
}
for(int i=0; i<k; ++i)
cnt[Index[i]]=0;
k=0;
int ans=0;
for(int x1=1; x1<=100; ++x1)
for(int x2=1; x2<=100; ++x2)
cnt[Index[k++]=-(a*x1*x1+b*x2*x2)+1000000]++;
for(int x3=1; x3<=100; ++x3)
for(int x4=1; x4<=100; ++x4)
ans+=cnt[c*x3*x3+d*x4*x4+1000000];
printf("%d\n",ans*16);
}
return 0;
}解法五(46ms)：用哈希表来记录cnt值，节省了大量空间，而且查找速度也较map快了许多。其余同三、四。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#define FRER() freopen("i.txt","r",stdin)
#define FREW() freopen("o.txt","w",stdout)
using namespace std;
typedef long long LL;
const int N=20000+100;
const int mod=(1<<15)-1;
int head[mod],nxt[N],num[N],cnt[N],nmem;
int Hash(int x)
{
return (x%mod+mod)%mod;
}
void Insert(int x)
{
int H=Hash(x);
for(int u=head[H]; ~u; u=nxt[u])
{
if(num[u]==x)
{
cnt[u]++;
return;
}
}
nxt[nmem]=head[H],num[nmem]=x,cnt[nmem]=1,head[H]=nmem++;
}
int Search(int x)
{
int H=Hash(x);
for(int u=head[H]; ~u; u=nxt[u])
{
if(num[u]==x)
return cnt[u];
}
return 0;
}
int main()
{
//FRER();
int a,b,c,d;
while(scanf("%d%d%d%d",&a,&b,&c,&d)==4)
{
if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
{
printf("0\n");
continue;
}
memset(head,-1,sizeof head);
nmem=0;
int ans=0;
for(int x1=1; x1<=100; ++x1)
for(int x2=1; x2<=100; ++x2)
Insert(-(a*x1*x1+b*x2*x2));
for(int x3=1; x3<=100; ++x3)
for(int x4=1; x4<=100; ++x4)
ans+=Search(c*x3*x3+d*x4*x4);
printf("%d\n",ans*16);
}
return 0;
}在vj上，如果不加全正或全负的判断的话，只有最后两种解法能过...