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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 |
裸的暴力题，比赛的时候开了5个循环，看时间限制在5秒，以为呀，结果本地都跑不过来。
第一次用short的说。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
short hash[25000001];
int a1,a2,a3,a4,a5;
int main()
{
while(~scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5))
{
int ans=0;
memset(hash,0,sizeof(hash));
for(int x1=-50;x1<=50;x1++)
{
if(x1==0)
continue;
for(int x2=-50;x2<=50;x2++)
{
if(x2==0)
continue;
int s=(-1)*(a1*x1*x1*x1+a2*x2*x2*x2);
if(s<0)
s+=25000000;
hash[s]++;
}
}
for(int x1=-50;x1<=50;x1++)
{
if(x1==0)
continue;
for(int x2=-50;x2<=50;x2++)
{
if(x2==0)
continue;
for(int x3=-50;x3<=50;x3++)
{
if(x3==0)
continue;
int s=a3*x1*x1*x1+a4*x2*x2*x2+a5*x3*x3*x3;
if(s<0)
s+=25000000;
if(hash[s])
ans+=hash[s];
}
}
}
printf("%d\n",ans);
}
return 0;
}
POJ 1840 Eqs（暴力）,布布扣,bubuko.com
时间： 2024-10-11 00:06:12