POJ 1840:Eqs 哈希求解五元方程
Eqs
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 14169 | Accepted: 6972 |
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.
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
给定方程是a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 ,给了a1,a2,a3,a4,a5。求解 解的数量。x都在-50到50之间。
暴搜肯定GG,就得把方程变形将a1x13+ a2x23 移到方程左边,作为结果,然后暴搜x3,x4,x5。时间复杂度就降下来了。
代码:
#include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <string> #include <cstring> #pragma warning(disable:4996) using namespace std; short hash1[25000000]; int main() { int a1, a2, a3, a4, a5; int x1, x2, x3, x4, x5; int temp; long long sum; scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5); memset(hash1, 0, sizeof(hash1)); for (x1 = -50; x1 <= 50; x1++) { if (x1 == 0)continue; for (x2 = -50; x2 <= 50; x2++) { if (x2 == 0)continue; temp = -1*(a1*x1*x1*x1 + a2*x2*x2*x2); if (temp < 0) { temp += 25000000; } hash1[temp]++; } } sum = 0; for (x3 = -50; x3 <= 50; x3++) { if (x3 == 0)continue; for (x4 = -50; x4 <= 50; x4++) { if (x4 == 0)continue; for (x5 = -50; x5 <= 50; x5++) { if (x5 == 0)continue; temp = a3*x3*x3*x3 + a4*x4*x4*x4 + a5*x5*x5*x5; if (temp < 0) { temp += 25000000; } if (temp >= 0 && temp < 25000000) { sum += hash1[temp]; } } } } cout << sum << endl; return 0; }
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