UVA 1397 - The Teacher's Side of Math(高斯消元)

UVA 1397 - The Teacher's Side of Math

题目链接

题意：给定一个x=a1/m+b1/n，求原方程组

思路：由于m*n最多20，所有最高项只有20，然后可以把每个此项拆分，之后得到n种不同无理数，每一项为0，就可以设系数为变元，构造方程进行高斯消元

一开始用longlong爆了，换成分数写法也爆了，又不想改高精度，最后是机智的用了double型过的，不过用double精度问题，所以高斯消元的姿势要正确，并且最后输出要注意-0的情况

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

typedef long long ll;

const int N = 25;
const double eps = 1e-9;

ll a, m, b, n, C

;
int hash

, tot;

double A

;

void build() {
memset(A, 0, sizeof(A));
A[0][0] = 1;
for (int i = 1; i <= tot; i++) {
for (int j = 0; j <= i; j++) {
int l = j, r = i - j;
double tmp = C[i][l] * pow(a * 1.0, l / m) * pow(b * 1.0, r / n);
l %= m; r %= n;
A[hash[l][r]][i] += tmp;
}
}
A[tot][tot] = 1;
A[tot][tot + 1] = 1;
tot++;
}

void getC() {
for (int i = 0; i <= 20; i++) {
C[i][0] = C[i][i] = 1;
for (int j = 1; j < i; j++)
C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
}

void gethash() {
tot = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
hash[i][j] = tot++;
}
}
}

void print(double x) {
char s[100];
sprintf(s, "%.0lf", x);
if (strcmp(s, "-0") == 0) printf(" %s", s + 1);
else printf(" %s", s);
}

void gauss() {
for (int i = 0; i < tot; i++) {
int r = i;
for (int j = i + 1; j < tot; j++) {
if (fabs(A[j][i]) > fabs(A[r][i]))
r = j;
}
if (fabs(A[r][i]) < eps) continue;
for (int j = i; j <= tot; j++)
swap(A[r][j], A[i][j]);
for (int j = 0; j < tot; j++) {
if (i == j) continue;
if (fabs(A[j][i]) >= eps) {
double tmp = A[j][i] / A[i][i];
for (int k = i; k <= tot; k++)
A[j][k] -= tmp * A[i][k];
}
}
}
printf("1");
for (int i = tot - 2; i >= 0; i--)
print(A[i][tot] / A[i][i]);
printf("\n");
}

int main() {
getC();
while (~scanf("%lld%lld%lld%lld", &a, &m, &b, &n)) {
if (!a && !m && !b && !n) break;
gethash();
build();
gauss();
}
return 0;
}