计算二次方程ax ^ 2 + bx + c = 0的根。由以下公式给出：(-b + square_root(b2 - 4ac))/2a和(-b - square_root(b2 - 4ac))/2a

这里给出了无限级数展开的公式：sin x = x - x3/3! + x5/5! - x7/7! + ...这里，x是弧度，它取值范围-1 = x = 1。

#include stdio.h#include stdlib.h#include math.hint main(){int p, intCircle = 0, intSquare, intToss = 3000, intRM;

#include stdio.h#include limits.h// integer limits#include float.h// floating-point limits

#include stdio.hint main() {int a, b;a = 11;b = 99;// to take values from user input uncomment the below lines −

#include stdio.hint main() {int a, b, c;a = 11;b = 22;c = 33;if ( a b a c )

#include stdio.hint main() {int n,i,j;n = 5;for(i = 1; i = n; i++) {for(j = 1; j = i; j++)

#include stdio.hint main() {int n,i,j;n = 5;// number of rows.for(i = 1; i = n; i++) {for(j = 1; j = n-i; j++)

// -------------------------------------#include stdio.hint main() {int i;for(i = 1; i = 10; i++) {

#include stdio.hint main () {int ret;char oldname[] = file.txt;char newname[] = newfile.txt;

// #include stdio.h#include string.hint main () {char dest[20];char src[20];int len;strcpy(src, Yiibai Point);

// ---------------------------------------#include stdio.hint main() {int i, start, end;start = 1;

// -------------------------------------#include stdio.hint main() {char ch[5] = {H, e, l, l, o};

#include stdio.hint main() {char ch[5] = Hello;int i = 0;/* Print as a complete string */

#include stdio.hint main() {int set1[5] = {10, 20, 30, 40, 50};int set2[5] = {101, 201, 301, 401, 501};

// -----------------------#include stdio.hint main() {int a;int b;a = 10;b = 20;printf( Value of a = %d and value of b = %d\\n, a, b );

// #include stdio.hint main() {charch1;charch2;charch3;charch4;ch1 = \\t;ch2 = \\n;

// #include stdio.h#include math.hint main() {shorts;inti;longl;floatf;double d;printf( sin(s): %f\\n, sin(10));

#include stdio.hint main() {intnumber1;intnumber2;intnumber3;intnumber4;intnumber5;number1 = 10;

#include stdio.h#include wchar.hint main(void){wchar_t value = L\\u00f6;wchar_t wch = LI;