a²＋a³＝392 怎么解？ 第1页

       #include <stdio.h> #include <stdlib.h> #include <string.h> #include "math.h"  typedef struct {     float coef[8];  // 各项系数     int hi; // 最高次 } polyn_t;  // 初始化多项式 void polyn_init(polyn_t* p, char* const tokens[], int count) {     int n = 0;     while(count > 0) {         count--;         p->coef[n] = strtod(tokens[count], NULL);         n++;     }     p->hi = n - 1; }  // 多项式求值 float polyn_calc(polyn_t* p, float x) {     float f = 0;     for(int i = p->hi; i >= 0; i--) {         f *= x;         f += p->coef[i];     }     return f; }  // 显示多项式 void polyn_view(polyn_t* p) {     for(int i = p->hi; i >= 0; i--) {         if(i >= 2)             printf("%.3f * x^%d + ", p->coef[i], i);         else if(i == 1)             printf("%.3f * x + ", p->coef[i]);         else             printf("%.3f", p->coef[i]);     }     printf("
"); }  // 多项式除法 void polyn_div(polyn_t* p1, polyn_t* p2) {     if(p1->hi < p2->hi)         return;      polyn_t tmp, quot;      quot.hi = p1->hi - p2->hi;     while(p1->hi >= p2->hi) {         memset(&tmp, '', sizeof(tmp));         tmp.hi = p2->hi;         float n = p1->coef[p1->hi] / p2->coef[p2->hi];         int k = p1->hi - p2->hi;         for(int i = p2->hi; i >= 0; i--) {             tmp.coef[i + k] = p2->coef[i] * n;         }         for(int i = p1->hi; i >= 0; i--) {             p1->coef[i] -= tmp.coef[i];         }         quot.coef[k] = n;         p1->hi--;     }     memset(p1, '', sizeof(*p1));     memcpy(p1, &quot, sizeof(*p1)); }  // 在(x1, x2)区间内寻找一根 float polyn_searchroot(polyn_t* p, float x1, float x2) {     while(x2 - x1 > 1e-20) {         float mid = (x2 + x1) / 2;         float y1 = polyn_calc(p, x1);         float y2 = polyn_calc(p, x2);         float ymid = polyn_calc(p, mid);         if(ymid == 0) {             return mid;         }         if(y1 * ymid < 0)             x2 = mid;         else if(ymid * y2 < 0)             x1 = mid;         else             return FP_SNAN;     }     return FP_SNAN; }  // 解方程 void polyn_solve(polyn_t* p) {     int count = 1;     while(p->hi >= 1) {         if(polyn_calc(p, 0) == 0) {             polyn_div(p, &(polyn_t) { {0, 1}, 1});             printf("x%d = 0
", count);             count++;         }         else {             float y0 = polyn_calc(p, 0);             float x = x, y;             int state = 0;             for(float xi = 1e-10; xi < 1e10; xi *= 1.01) {                 y = polyn_calc(p, -xi);                 if(y * y0 < 0) {                     x = polyn_searchroot(p, -xi, 0);                     state = 1;                     break;                 }                 y = polyn_calc(p, xi);                 if(y * y0 < 0) {                     x = polyn_searchroot(p, 0, xi);                     state = 1;                     break;                 }             }             if(state == 0) { //                printf("Can't find more roots.
");                 exit(1);             }             polyn_div(p, &(polyn_t) { {-x, 1}, 1});             printf("x%d = %f
", count, x);             count++;         }     } }  int main(void) {     static char buf[256], seps[] = ", ";     char* token;     char* tokens[8];     int count = 0;     fgets(buf, 256, stdin);      token = strtok(buf, seps);     while(token != NULL) {         tokens[count] = token;         token = strtok(NULL, seps); // Get next token:         count++;         if(count >= 8)             break;     }     if(count <= 0) {         exit(1);     }     polyn_t p;     polyn_init(&p, tokens, count);      polyn_view(&p);     polyn_solve(&p); }      

       1 -4 -12 1.000 * x^2 + -4.000 * x + -12.000 x1 = -2.000000 x2 = 6.000000     

       1 6 11 6 1.000 * x^3 + 6.000 * x^2 + 11.000 * x + 6.000 x1 = -1.000000 x2 = -2.000000 x3 = -3.000000     

       1 -10 35 -50 24 1.000 * x^4 + -10.000 * x^3 + 35.000 * x^2 + -50.000 * x + 24.000 x1 = 1.000000 x2 = 2.000000 x3 = 3.000000 x4 = 4.000000     

       1 -841 -167660 8568500 91600000 -100000000 1.000 * x^5 + -841.000 * x^4 + -167660.000 * x^3 + 8568500.000 * x^2 + 91600000.000 * x + -100000000.000 x1 = 1.000000 x2 = -10.000000 x3 = 50.000000 x4 = -200.000000 x5 = 1000.000000