Eqs（sort+二分）

I - Eqs

Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1840

Appoint description: System Crawler  (2015-10-03)

Description

Consider equations having the following form: 
a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=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

   这题是简单的二分题，没什么难度的。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<cstdio>#include<vector>#include<cmath>using namespace std;#define T 105#define inf 0x3f3f3f3f#define CRL(a) memset(a,inf,sizeof(a))typedef long long ll;int a1,a2,a3,a4,a5;vector<int> n,m;int main(){/*freopen("input.txt","r",stdin);*/int x1,x2,x3,x4,cnt,c=0;while(~scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5)){cnt=0;n.clear(),m.clear();for(x1=-50;x1<=50;++x1)if(x1)for(x2=-50;x2<=50;++x2)if(x2)for(x3=-50;x3<=50;++x3)if(x3){n.push_back(a1*x1*x1*x1+a2*x2*x2*x2+a3*x3*x3*x3);}for(x4=-50;x4<=50;++x4)if(x4)for(int x5=-50;x5<=50;++x5){if(x5)m.push_back(a4*x4*x4*x4+a5*x5*x5*x5);}sort(n.begin(),n.end());for(int i=0;i<m.size();++i){cnt+=upper_bound(n.begin(),n.end(),-m[i])-lower_bound(n.begin(),n.end(),-m[i]);}printf("%d\n",cnt);}return 0;}

网上大牛代码（hash）：

#include <stdio.h>#include <string.h>short hash[25000001];int main(){    int a1,a2,a3,a4,a5,x1,x2,x3,x4,x5,sum;    while(~scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5))    {        memset(hash,0,sizeof(hash));        for(x1 = -50; x1<=50; x1++)        {            if(!x1)                continue;            for(x2 = -50; x2<=50; x2++)            {                if(!x2)                    continue;                sum = -1*(a1*x1*x1*x1+a2*x2*x2*x2);                if(sum<0)                    sum+=25000000;//数字下标没有负数，换一种出储存方式                hash[sum]++;            }        }        int cnt = 0;        for(x3 = -50; x3<=50; x3++)        {            if(!x3)                continue;            for(x4 = -50; x4<=50; x4++)            {                if(!x4)                    continue;                for(x5 = -50; x5<=50; x5++)                {                    if(!x5)                        continue;                    sum = a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5;                    if(sum<0)                        sum+=25000000;                    cnt+=hash[sum];                }            }        }        printf("%d\n",cnt);    }    return 0;}