POJ训练计划1840_Eqs(二分)

Eqs

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 11729 Accepted: 5744

Description

Consider equations having the following form: 
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

The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.

Output

The output will contain on the first line the number of the solutions for the given equation.

Sample Input

37 29 41 43 47

Sample Output

654

Source

Romania OI 2002

解题报告

100^5的时间复杂度肯定挂掉的，想只用100^4+二分看能不能过，结果还是不行。。。

空间换时间的概念存下a1和a2的所有情况。在100^3和二分情况下还要考虑存下的情况有重复答案，都要计算。。。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int num[10010];int main(){    int a[6],i,j,k,l,r,n,m,cnt=0,mid,t;    for(i=0;i<5;i++)        scanf("%d",&a[i]);    n=0;    for(i=-50;i<=50;i++)    {        if(i)        {            for(j=-50;j<=50;j++)            {                if(j)                {                    num[n++]=a[0]*i*i*i+a[1]*j*j*j;                }            }        }    }    sort(num,num+n);    for(i=-50;i<=50;i++)    {        if(i)        {            for(j=-50;j<=50;j++)            {                if(j)                {                    for(k=-50;k<=50;k++)                    {                        if(k)                        {                            int key=-(a[2]*i*i*i+a[3]*j*j*j+a[4]*k*k*k);                            l=0,r=n-1;                            while(l<=r)                            {                                mid=(l+r)/2;                                if(num[mid]<key)                                    l=mid+1;                                else if(num[mid]>key)                                    r=mid-1;                                else                                {                                    cnt++;                                    break;                                }                            }                            for(t=mid+1;t<n&&num[t]==key;t++)                                cnt++;                            for(t=mid-1;t>=0&&num[t]==key;t--)                                cnt++;                        }                    }                }            }        }    }    printf("%d\n",cnt);    return 0;}