题解 P1022 【计算器的改良 】
xzqiaochu
2018-10-26 00:13:25
第一步：把输入的方程转化成 f(x) = 0 的形式
第二步：用二分法逼近求该函数的近似根
代码如下：
```
#include <iostream>
#include <cstring>
#include <math.h>
#define maxn 100
#define inf 1000
using namespace std;
double f(char bds[], double x);
char x_sign = 0;
int main() {
cout.setf(ios::fixed);
cout.precision(3);
char bds_input[maxn];
char * bds1;
char * bds2;
cin >> bds_input;
bds1 = strtok(bds_input, "=#");
bds2 = strtok(NULL, "=#");
double x1 = -inf, x2 = inf, xm = 0;
while(1) {
double y1 = f(bds1, x1) - f(bds2, x1);
double y2 = f(bds1, x2) - f(bds2, x2);
xm = (x1 + x2) / 2;
if (fabs(x1 - x2) <= 0.0001) break;
double ym = f(bds1, xm) - f(bds2, xm);
if (y2 * ym <= 0) {
x1 = xm;
}else{
x2 = xm;
}
}
cout << x_sign << '=' << xm;
return 0;
}
double f(char bds[], double x) {
bool sign = true;
double sum = 0, temp = 0;
int i = 0;
if (bds[i] == '-') {
sign = false;
i++;
}
for (; i <= strlen(bds) - 1; i++) {
if (bds[i] == '+') {
sign ? sum += temp : sum -= temp;
temp = 0;
sign = true;
continue;
}
if (bds[i] == '-') {
sign ? sum += temp : sum -= temp;
temp = 0;
sign = false;;
continue;
}
if (isalpha(bds[i])) {
temp == 0 ? temp = x : temp *= x;
x_sign = bds[i];
}
if (bds[i] >= 48 && bds[i] <= 57) {
temp = temp * 10 + bds[i] - 48;
}
if (i == strlen(bds) - 1) {
sign ? sum += temp : sum -= temp;
}
}
return sum;
}
```