poj 1840(移動幾項到等式另外一邊減少循環深度)

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 11220		Accepted: 5480

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

Romania OI 2002

上來寫了一個四層for循環，結果直接tle了，優化了幾發還是tle。

將前三項移動到等式右面，就可以降低到O(n^2 + n^3)的複雜度：

-(a1x1³+ a2x2³+ a3x3³ )=  a4x4³+ a5x5³

把等式左面可能的取值放到一個數組中做hash，然後遍歷右面的取值即可。

提交記錄：

1,Wrong Answer 沒有看到題目中的xi！=0 這個條件。

2.TLE 四層循環直接掛掉。。

3.Accepted 修改爲三等循環+兩層循環

代碼：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define oo 0x3f3f3f3f
#define MAX_N 30010
#define MAX_M 510
using namespace std;
int a1, a2, a3, a4, a5;
int x1, x2, x3, x4, x5;
int data[50*50*50*10] = {0};
int data1[60] = {0};
short int hash[50*50*50*50*4 + 100] = {0};
map m;
int main() {
    int i;
    for (i = 0; i <= 50; i++) {
        data[i*i*i] = i;
        data1[i] = i*i*i;
    }
    vector::iterator it;
    int mod = 50*50*50*50*2;
    while (EOF != scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5)) {
        int result = 0, tmp;
        memset(hash, 0, sizeof(hash));
        for (x1 = -50; x1 <= 50; x1++) {
            int x13;
            if (x1 < 0) x13 = - data1[-x1] * a1;
            else x13 = data1[x1] * a1;
            if (x1 != 0)
            for (x2 = -50; x2 <= 50; x2++) {
                int x23;
                if (x2 < 0) x23 = - data1[-x2] * a2;
                else x23 = data1[x2] * a2;
                if (x2 != 0)
                for (x3 = -50; x3 <= 50; x3++) {
                    int x33;
                    if (x3 < 0) x33 = - data1[-x3] * a3;
                    else x33 = data1[x3] * a3;
                    if (x3 != 0) {
                        tmp = x13 + x23 + x33;
                        tmp = -tmp;
                        if (tmp < mod && tmp+mod >= 0) 
                        hash[tmp+mod]++;
                        //if (m.find(tmp) != m.end()) m[tmp]++;
                        //else m[tmp] = 1;
                    }
                }
            }
        }
        for (x4 = -50; x4 <= 50; x4++) {
            if (x4 != 0) {
                int x43;
                if (x4 < 0) x43 = -data1[-x4] * a4;
                else x43 = data1[x4] * a4;
                for (x5 = -50; x5 <= 50; x5++) {
                    if (!x5) continue;//排除了那種特殊情況
                    int x53;
                    if (x5 < 0) x53 = -data1[-x5] * a5;
                    else x53 = data1[x5] * a5;
                    tmp = x43 + x53;
                    result += hash[tmp+mod];
                    //if (m.find(tmp) != m.end()) result += m[tmp];
                }
            }
        }
        cout << result << endl;
    }
    return 0;
}