您好我正在研究计算器，我可以选择要解决的问题。首先，你会被问到你有多少。对于这段代码，我刚给出了3个。因此，默认情况下，输入3.接下来，我有一个11的列表给出，以解决所有方程。用户将从1-11中选择3个数字...我已经使用数字1,3,9和1,3,10以及1,3,11开始了一个案例。所以我的问题是，当我随机选择1,3,10时，例如我通过选择3,10,1而不是1,3,10来改变我在输入中的顺序，选择2，选择3 ...... / p>
所以我想说我会选择1,3,10但是我按照这个顺序输入了它... 10,3,1 ..但仍然必须去实施它将要实施的动作。
我在if语句中使用了以下行... if（选择== 1,3,10＆amp; choice2 == 1,3,10＆amp; 7 choice3 == 1,3,10）然后执行行动......
和
if（（choice == 1 || 3 || 10）＆amp;＆amp;（choice2 == 1 || 3 || 10）＆amp;＆amp;（choice3 == 1 || 3 || 10））然后执行行动...
我已尝试过上述内容，但它不会执行下面的语句......但它会执行上面的语句......
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define PI 3.14159265
int main(){
double length, angle, radius, tangent, chord, midordinate, external, degree;
double pcurve,ptan,pintersect;
double ulength, udegree, uangle, uradius, utangent, uchord, umidordinate, uexternal;
int choice, choice2, choice3, given;
//For sin, cos, tan
double x, ret, val;
val = PI / 180;
printf("Enter number of given: ");
scanf("%d",&given);
if(given==3){
choice:
printf("[1] - Angle\n");
printf("[2] - Degree\n");
printf("[3] - Radius\n");
printf("[4] - Length of Curve\n");
printf("[5] - Tangent\n");
printf("[6] - Chord\n");
printf("[7] - Midordinate\n");
printf("[8] - External Distance\n");
printf("[9] - Point of Intersection\n");
printf("[10] - Point of Curve\n");
printf("[11] - Point of Tangent\n");
printf("\n");
printf("Enter 1st given: ");
scanf("%d",&choice);
printf("Enter 2nd given: ");
scanf("%d",&choice2);
printf("Enter 3rd given: ");
scanf("%d",&choice3);
printf("\n-----------------------------------\n");
if(choice==1,3,9 && choice2==1,3,9 && choice3==1,3,9){
printf("Enter angle: ");
scanf("%lf",&angle);
printf("Enter radius: ");
scanf("%lf",&radius);
printf("Enter Point of Intersection (Point of Curve Value): ");
scanf("%lf",&pcurve);
printf("Enter Point of Intersection (Tangent Value): ");
scanf("%lf",&tangent);
printf("-----------------------------------\n");
printf("\nGIVEN:\n");
printf("-----------------------------------\n");
printf("Angle = %lf\n",angle);
printf("Radius = %lf\n",radius);
printf("Point of Intersection (PI) = %lf + %lf\n", pcurve,tangent);
uangle = angle/2;
printf("-----------------------------------\n");
printf("\nRESULTS:\n");
printf("-----------------------------------\n");
length=(radius*angle*PI)/180;
tangent = radius * (tan(uangle*val));
chord = 2*radius*(sin(uangle*val));
midordinate = radius - (radius*(cos(uangle*val)));
external = radius *( (1/(cos (uangle*val) ) ) - 1 ) ;
pintersect = pcurve + tangent;
pcurve = pintersect - tangent;
ptan = pcurve + length;
printf("Radius = %lf\n",radius);
printf("Length of Curve = %lf\n",length);
printf("Tangent = %lf\n",tangent);
printf("Chord = %lf\n",chord);
printf("Mid Ordinate = %lf\n",midordinate);
printf("External Distance = %lf\n",external);
printf("Point of Intersection = %lf\n",pintersect);
printf("Point of Curve = %lf\n",pcurve);
printf("Point of Tangent = %lf\n",ptan);
}
else
if( (choice== 1 || 3 || 10) && (choice2== 1 || 3 || 10) && (choice3== 1 || 3 || 10)){
printf("Enter angle: ");
scanf("%lf",&angle);
printf("Enter radius: ");
scanf("%lf",&radius);
printf("Enter Point of Curve (Point of Intersection Value): ");
scanf("%lf",&pintersect);
printf("Enter Point of Curve (Tangent Value): ");
scanf("%lf",&tangent);
printf("-----------------------------------\n");
printf("\nGIVEN:\n");
printf("-----------------------------------\n");
printf("Angle = %lf\n",angle);
printf("Radius = %lf\n",radius);
printf("Point of Curve (PC) = %lf - %lf\n", pintersect,tangent);
uangle = angle/2;
printf("-----------------------------------\n");
printf("\nRESULTS:\n");
printf("-----------------------------------\n");
length=(radius*angle*PI)/180;
tangent = radius * (tan(uangle*val));
chord = 2*radius*(sin(uangle*val));
midordinate = radius - (radius*(cos(uangle*val)));
external = radius *( (1/(cos (uangle*val) ) ) - 1 ) ;
pintersect = pcurve + tangent;
pcurve = pintersect - tangent;
ptan = pcurve + length;
printf("Radius = %lf\n",radius);
printf("Length of Curve = %lf\n",length);
printf("Tangent = %lf\n",tangent);
printf("Chord = %lf\n",chord);
printf("Mid Ordinate = %lf\n",midordinate);
printf("External Distance = %lf\n",external);
printf("Point of Intersection = %lf\n",pintersect);
printf("Point of Curve = %lf\n",pcurve);
printf("Point of Tangent = %lf\n",ptan);
}
if(choice==1,3,11 && choice2==1,3,11 && choice3==1,3,11){
printf("Enter angle: ");
scanf("%lf",&angle);
printf("Enter radius: ");
scanf("%lf",&radius);
printf("Enter Point of Tangent (Point of Curve Value): ");
scanf("%lf",&pintersect);
printf("Enter Point of Tangent (Length Value): ");
scanf("%lf",&length);
printf("-----------------------------------\n");
printf("\nGIVEN:\n");
printf("-----------------------------------\n");
printf("Angle = %lf\n",angle);
printf("Radius = %lf\n",radius);
printf("Point of Curve (PC) = %lf + %lf\n", pcurve,length);
uangle = angle/2;
printf("-----------------------------------\n");
printf("\nRESULTS:\n");
printf("-----------------------------------\n");
length=(radius*angle*PI)/180;
tangent = radius * (tan(uangle*val));
chord = 2*radius*(sin(uangle*val));
midordinate = radius - (radius*(cos(uangle*val)));
external = radius *( (1/(cos (uangle*val) ) ) - 1 ) ;
pintersect = pcurve + tangent;
pcurve = pintersect - tangent;
ptan = pcurve + length;
printf("Radius = %lf\n",radius);
printf("Length of Curve = %lf\n",length);
printf("Tangent = %lf\n",tangent);
printf("Chord = %lf\n",chord);
printf("Mid Ordinate = %lf\n",midordinate);
printf("External Distance = %lf\n",external);
printf("Point of Intersection = %lf\n",pintersect);
printf("Point of Curve = %lf\n",pcurve);
printf("Point of Tangent = %lf\n",ptan);
}
}
getch();
return 0;
}
答案 0 :(得分：1)
if(choice==1,3,9 && choice2==1,3,9 && choice3==1,3,9){
if( (choice== 1 || 3 || 10) && (choice2== 1 || 3 || 10) && (choice3== 1 || 3 || 10)) //always TRUE
if(choice==1,3,11 && choice2==1,3,11 && choice3==1,3,11){
您是否认为这些条件是符合您需求的有效语法？
正确
if((choice==1 ||choice==3 ||choice==9) && (choice2==1 ||choice2==3 ||choice2==9) && (choice3==1 ||choice3==3 ||choice3==9))
if((choice==1 ||choice==3 ||choice==10) && (choice2==1 ||choice2==3 ||choice2==10) && (choice3==1 ||choice3==3 ||choice3==10))
if((choice==1 ||choice==3 ||choice==11) && (choice2==1 ||choice2==3 ||choice2==11) && (choice3==1 ||choice3==3 ||choice3==11))