322 lines
7.9 KiB
C
322 lines
7.9 KiB
C
#include <stdio.h>
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#include <stdlib.h>
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#include <math.h>
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#include <string.h>
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#define MAX 10
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// 儲存每一項的結構
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struct Term {
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int coefficient; // 係數
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int exponent; // 指數
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};
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//計算可能的解
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void count_root(int arrR[],int arrL[],int ans[],int countF,int countL ,int *len_ans){
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int i , j ,k=0;
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for(i=0;i<countF;i++){
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for(j=0;j<countL;j++){
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if((arrL[j]%arrR[i])==0){
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ans[k++] = arrL[j]/arrR[i];
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}
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}
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}*len_ans =k;
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}
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//將最高次係數與常數的因數加入負號
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void plus_root_negative(int arr[],int *len){
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int i , j=0;
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for(i=0;i<*len;i++){
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arr[*len+j]=-(arr[i]);
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j++;
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}*len=*len+j;
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}
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//找最高次係數與常數的因數
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void find_factor(int P[], int facF[], int facL[], int len, int *count1, int *count2) {
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int i, j = 0;
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for (i = 1; i <= abs(ceil(P[2] / 2)); i++) {
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if (abs(P[2]) % i == 0) {
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facF[j++] = i;
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}
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}
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facF[j++] = fabs(P[2]); // 包含 P[2] 本身
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*count1 = j; // 設定找到的因數個數
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j = 0;
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for (i = 1 ; i <= abs(ceil(P[len-1] / 2)); i++) {
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if (abs(P[len-1]) % i == 0) {
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facL[j++] = i;
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}
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}
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facL[j++] = fabs(P[len-1]); // 包含 P[len] 本身
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*count2 = j; // 設定找到的因數個數
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}
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//把根代入方程式
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int evaluate_polynomial(int A[], int len, int x) {
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int result = 0;
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int i;
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for (i = 2; i < len; i += 2) {
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result += A[i] * pow(x, A[i-1]);
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}
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return result;
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}
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//把根代入方程式 求答案
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void Substitute(int len_root, int arr[], int ans[], int A[], int len_A, int *len_ans) {
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int i, j = 0, result;
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for (i = 0; i < len_root; i++) {
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result = evaluate_polynomial(A, len_A, arr[i]);
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// printf("x = %d, result = %d\n", arr[i], result);
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if (result == 0) {
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ans[j++] = arr[i];
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}
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}
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*len_ans = j;
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}
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//移除重複的答案
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void remove_duplicates(int A[], int len, int new_A[], int *new_len) {
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int i, j, k;
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int is_duplicate;
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*new_len = 0; // 初始化新的陣列長度為0
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// 遍歷原始陣列
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for (i = 0; i < len; i++) {
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is_duplicate = 0; // 假設當前元素沒有重複
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// 檢查是否在結果陣列中出現過
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for (j = 0; j < *new_len; j++) {
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if (A[i] == new_A[j]) {
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is_duplicate = 1; // 如果發現重複，標記為1
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break;
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}
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}
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// 如果沒有重複，將其加入結果陣列
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if (!is_duplicate) {
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new_A[*new_len] = A[i];
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(*new_len)++; // 更新結果陣列的長度
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}
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}
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}
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//顯示方程式
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void output(int *A,int len){
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int i ,j,flag_Max=0,flag_x=0, flag_const=0 ;
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printf("\n方程式p(x)= ");
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//A[i]為指數 A[i+1]為係數
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//i從1開始數 一次讀取一個指數與係數
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//len為A[]的長度
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//flag_Max 為最高次項的旗標
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//flag_x 為1次項的旗標
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//flag_const 為常數項的旗標
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for(i=1;i<len;i++){
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if(A[i]==1){ //若指數項=1 觸發一次項旗標
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flag_x=1;
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}else if(A[i]==0){ //若指數項=0 觸發常數項旗標
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flag_const=1;
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}
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if(flag_Max==0){ //第一項
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if( A[i]>1 && A[i+1]!=1){ //指數大於1 係數不等於1
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printf("%dx^%d",A[i+1],A[i]);
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}else if(A[i]>1 && A[i+1]==1){
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printf("x^%d",A[i]); //指數大於1 係數等於1
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}else if(A[i]==1 && A[i+1]!=1){ //指數=1 係數不等於1
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printf("%dx",A[i+1]);
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}else if(A[i]==0 && A[i+1]!=1){ //指數=0 係數不等於1
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printf("%d",A[i+1]);
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}else if(A[i]==1 && A[i+1]==1){ //指數=1 係數=1
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printf("x");
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}else if(A[i]==1 && A[i+1]==-1){ //指數=1 係數=-1
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printf("-x");
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}flag_Max = 1; //高次項的旗標=1
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}else if(flag_x == 1 && flag_const==0){// x一次項
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if(A[i+1]>0){ // x一次項為正
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if(A[i+1]==1){
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printf("+x",A[i+1]);
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}else{
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printf("+%dx",A[i+1]);
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}
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}else if(A[i+1]<0){ //x一次項為負
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if(A[i+1]==-1){
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printf("-x",A[i+1]);
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}else{
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printf("%dx",A[i+1]);
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}
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}
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}else if(flag_const==1){ // 常數項
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if(A[i+1]>0){ //常數項>0
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printf("+%d",A[i+1]);
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}else if(A[i+1]<0) { //常數項<0
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printf("%d",A[i+1]);
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}
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}else if(flag_Max==1 && flag_x==0 && flag_const==0) { //多項式中間的部分
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if(A[i+1]>0){ //係數>1
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if(A[i+1]==1){ //係數=1
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printf("x^%d",A[i]);
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}else{ //係數不等於1
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printf("+%dx^%d",A[i+1],A[i]);
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}
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}else if(A[i+1]<0){ //係數<1
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if(A[i+1]==-1){ //係數=-1
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printf("-x^%d",A[i]);
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}else{ //係數不等於1
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printf("%dx^%d",A[i+1],A[i]);
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}
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}
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}i++;
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}
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printf("\n");
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}
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int main() {
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int i,j, len, countF, countL, len_root, len_ans ,len_new ;
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char input[100];
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struct Term terms[MAX];
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int term_count = 0;
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printf("\n*** 解多項式之整數根 ***\n");
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// 1. 讀取輸入
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printf("\n請輸入多項式： ");
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fgets(input, sizeof(input), stdin);
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input[strcspn(input, "\n")] = 0; // 移除換行符
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// 2. 處理每一項
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char *ptr = input;
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while (*ptr && term_count < MAX) {
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int coef = 0, exp = 0;
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int is_negative = 0;
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// 檢查負號
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if (*ptr == '-') {
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is_negative = 1;
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ptr++;
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} else if (*ptr == '+') {
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ptr++;
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}
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// 跳過空白字符
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while (isspace(*ptr)) ptr++;
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// 3. 解析每一項
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if (sscanf(ptr, "%dx^%d", &coef, &exp) == 2) {
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// 完整的 ax^b 形式
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ptr = strchr(ptr, '^') + 1;
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} else if (sscanf(ptr, "%dx", &coef) == 1) {
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// ax 形式，指數為1
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exp = 1;
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ptr = strchr(ptr, 'x') + 1;
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} else if (sscanf(ptr, "x^%d", &exp) == 1) {
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// x^b 形式，係數為1
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coef = 1;
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ptr = strchr(ptr, '^') + 1;
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} else if (strncmp(ptr, "x", 1) == 0) {
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// 單獨的 x，係數為1，指數為1
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coef = 1;
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exp = 1;
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ptr++;
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} else {
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// 假設是常數項
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sscanf(ptr, "%d", &coef);
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exp = 0;
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while (isdigit(*ptr)) ptr++;
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}
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// 4. 處理負號
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if (is_negative) {
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coef = -coef;
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}
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// 5. 儲存結果
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terms[term_count].coefficient = coef;
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terms[term_count].exponent = exp;
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term_count++;
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// 輸出當前項的信息（用於調試）
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// printf("項 %d: 係數 = %d, 指數 = %d\n", term_count, coef, exp);
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// 移動到下一個 '+' 或 '-' 符號
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while (*ptr && *ptr != '+' && *ptr != '-') ptr++;
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}
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// int *A = (int*)malloc(((term_count*2)+1) * sizeof(int)) ;
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len = ((term_count*2)+1);
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int A[len];
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// printf("term_count=%d\n",term_count);
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// printf("((term_count*2)+1)=%d\n",((term_count*2)+1));
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// printf("sizeof(A)=%d\n",sizeof(A));
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A[0]=term_count;
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for (i = 1 , j=0 ; i < term_count*2 ; i++ , j++ ){
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A[i]= terms[j].exponent;
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A[i+1]= terms[j].coefficient;
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// printf("A[%d]=%d\n",i,terms[j].exponent);
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// printf("A[%d]=%d\n",i+1,terms[j].coefficient);
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i++;
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}
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// printf("A[]= ");
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// for (i = 0; i < len; i++) {
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// printf(" %d ",A[i]);
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// }
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// printf("\n");
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// 1x^3-2x^2-1x^1+2x^0
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// 1x^5-1x^3-1x^2-1x^1+1x^0
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// 1x^2-100x^0
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// 1x^2+100x^0
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// 3x^4+9x^3-198x^2+156x^1+360x^0
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int facF[30], facL[30], root[60], ans[60] ,new_ans[8];
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find_factor(A, facF, facL, len, &countF, &countL);
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// printf("最高次方項因數為: ");
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// for (i = 0; i < countF; i++) {
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// printf("%d ", facF[i]);
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// }
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// printf("\n常數項因數為: ");
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// for (i = 0; i < countL; i++) {
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// printf("%d ", facL[i]);
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// }
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// printf("\n");
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count_root(facF, facL, root, countF, countL, &len_root);
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plus_root_negative(root, &len_root);
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// printf("可能的根：\n");
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// for (i = 0; i < len_root; i++) {
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// printf("root[%d] = %d\n", i, root[i]);
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// }
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//輸出多項式
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output(A,len);
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// 計算可能的解
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Substitute(len_root, root, ans, A, len, &len_ans);
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//移除多餘的解
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remove_duplicates(ans,len_ans,new_ans,&len_new);
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if(len_new==0){
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printf("\n整數根不存在");
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return 1 ;
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}
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if(len_new>0){
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printf("\n解得整數根為：");
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for (i = 0; i < len_new; i++) {
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if(i<len_new-1) {
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printf("%d、",new_ans[i]);
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}else if(i==len_new-1) {
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printf("%d", new_ans[i]);
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}
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}
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}
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return 0;
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}
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