POJ 1840 Eqs

解题思路：hash+枚举

0 0 0 0 0 输出 1e10，因此需要64位整数

某一系数为0，做一点优化

#include <iostream>
using namespace std;
#define PRIME 100003
#define MAXN 1000001
#define Cube(n) ((n)*(n)*(n)) 
int p[MAXN],first[PRIME],next[MAXN];
int main()
{
    int i,j,a,b,c,h,t,coef[5],iter=1000000,zero=0;
    long long ans=0;
    for(i=0;i<PRIME;i++)first[i]=-1;
    for(i=0;i<5;i++)scanf("%d",&coef[i]);
    for(i=2;i>=0;i--)if(!coef[i]){iter/=100;zero++;}else if(zero){coef[i+zero]=coef[i];coef[i]=0;}
    for(j=i=0;i<iter;i++,j=i)
    {
        if(!coef[0])a=0;else{a=j/10000-50;if(a>=0)a++;j%=10000;}
        if(!coef[1])b=0;else{b=j/100-50;if(b>=0)b++;j%=100;}
        if(!coef[2])c=0;else{c=j-50;if(c>=0)c++;}
        h = coef[0]*Cube(a)+coef[1]*Cube(b)+coef[2]*Cube(c);
        p[i]=h;
        h%=PRIME;if(h<0)h+=PRIME;
        next[i]=first[h];first[h]=i;
    }
    for(iter=1e4,i=3;i<5;i++)if(!coef[i])iter/=100,zero++;
    if(!coef[4]){coef[4]=coef[3];coef[3]=0;}
    for(j=i=0;i<iter;i++,j=i)
    {
        if(!coef[3])a=0;else{a=j/100-50;if(a>=0)a++;}
        if(!coef[4])b=0;else{b=j%100-50;if(b>=0)b++;}
        c=-coef[3]*Cube(a)-coef[4]*Cube(b);
        h=c%PRIME;if(h<0)h+=PRIME;
        for (t=first[h];t!=-1;t=next[t])
            if(p[t]==c)ans++;
    }
    while(zero--)ans*=100;
    printf("%lld\n",ans);
    return 0;
}