Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 16229 | Accepted: 7965 |
Description
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0
The coefficients are given integers from the interval [-50,50].
It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}.
Determine how many solutions satisfy the given equation.
Input
Output
#include<iostream>
#include<cstring>
#include<string>
using namespace std;
short hashs[25000001];
int main()
{
int a,b,c,d,e;
while(cin>>a>>b>>c>>d>>e){
int sb=0;
memset(hashs,0,sizeof(hashs));
for(int i=-50;i<=50;i++)
{
if(i==0)
continue;
for(int j=-50;j<=50;j++)
{
if(j==0)
continue;
int sum=-(a*i*i*i+b*j*j*j);
if(sum<0)
sum+=25000000;
hashs[sum]++;
}
}
for(int i=-50;i<=50;i++)
{
if(i==0)
continue;
for(int j=-50;j<=50;j++)
{
if(j==0)
continue;
for(int k=-50;k<=50;k++)
{
if(k==0)
continue;
int sum=c*i*i*i+d*j*j*j+e*k*k*k;
if(sum<0)
sum+=25000000;
if(hashs[sum])
sb+=hashs[sum];
}
}
}
cout<<sb<<endl;
}
}
number of the solutions for the given equation.
Sample Input
37 29 41 43 47
Sample Output
654
Source
開始的思路中只有時間複雜度爲0(n^5)簡單做法,毫無疑問肯定要超時。沒有思路只好上別人家的博客做做客,原來是用到哈希!而且思路也不是很get到,爲啥自己想就那麼費勁,點播一下就開竅了呢?唉,什麼時候我能夠讓大把的人來我這做做客呢??還是要多練多思考!
推薦這篇博客http://blog.csdn.net/lyy289065406/article/details/6647387裏面講的非常詳細。值得學習代碼優化的思路