測試賽C - Eqs（雜湊）

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C - Eqs Time Limit:5000MS      Memory Limit:65536KB      64bit IO Format:%I64d & %I64uSubmit Status

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

雜湊存下一開始的兩個值，看找後面三個的和，看能不能出現0

讓5個for迴圈轉化為1個雙重for迴圈+1個三重for迴圈

#include <cstdio>#include <cstring>#include <algorithm>#define maxint 25000000using namespace std;short p[25000001];int main(){    int a, b, c, d, e, i, j, k, n, s;    while(scanf("%d %d %d %d %d", &a, &b, &c, &d, &e)!=EOF)    {        s = 0;        memset(p,0,sizeof(p));        for(i = -50; i <= 50 ; i++)        {            if(!i)                continue ;            for(j=-50; j<=50; j++)            {                if(!j)                    continue ;                n = i*i*i*a + j*j*j*b ;                n = -n ;                if(n<0)                    n += maxint ;                p[n]++;            }        }        for(i = -50 ; i <= 50; i++)        {            if(!i)                continue ;            for(j = -50; j <= 50; j++)            {                if(!j)                    continue ;                for(k = -50; k <= 50; k++)                {                    if(!k)                        continue ;                    n= i*i*i*c + j*j*j*d + k*k*k*e;                    if(n < 0)                        n += maxint ;                    if(p[n])                        s += p[n];                }            }        }        printf("%d\n",s);    }    return 0;}