角度（度分秒）的四则运算

角度（度分秒）的四则运算

在测量计算（如闭合导线坐标计算），我们可能需要计算比较多的角度。数学中，角度的度与分、分与秒之间一律采用六十进制，完全通过笔算耗时耗力，还容易算错。下面用C语言知识编写一个简单的程序，来实现角度的四则运算。

#include <stdio.h>
#include <stdlib.h>
typedef struct angle
{int degree;int minute;int second;
}ANGLE;
void decrease(ANGLE a[], int i);
void add(ANGLE a[], int i);
void multiply(ANGLE a[], int i, int j);
void divide(ANGLE a[], int i, int j);
int main()
{do{int i = 0, k, j = 0;float degree = 0;char operation[25];ANGLE a[25] = {0};printf("本程序可实现角度（度/分/秒）的四则运算，输入时角度的度、分、秒间以一个空格隔开\n");printf("角度 %d：",i+1);scanf("%d %d %d",&a[i].degree, &a[i].minute, &a[i].second);getchar();printf("运算符：");scanf("%c",&operation[i]);do{if(operation[i] == '+' || operation[i] == '-'){i++;printf("角度 %d：",i+1);scanf("%d %d %d",&a[i].degree, &a[i].minute, &a[i].second);getchar();printf("运算符：");scanf("%c",&operation[i]);}else if(operation[i] == '*'){printf("  乘数：");scanf("%d", &j);getchar();printf("运算符：");scanf("%c",&operation[i+1]);}else if(operation[i] == '/'){printf("  除数：");scanf("%d", &j);getchar();printf("运算符：");scanf("%c",&operation[i+1]);}}while(operation[i] != '=' && operation[i+1] != '=');for(k = 0; k < i+1; k++){if(operation[k] == '+'){add(a, k+1);}else if(operation[k] == '-'){decrease(a, k+1);}else if(operation[k] == '*'){multiply(a, k, j);}else if(operation[k] == '/'){divide(a, k, j);}else if(operation[k] != '-' && operation[k] != '+' && operation[k] != '=' && operation[k] != '/' && operation[k] != '*'){printf("请输入正确的运算符！\n");}}if(a[i].degree < 0 || a[i].minute < 0 || a[i].second < 0){printf("结果为：-%d°%d′%d″\n",abs(a[i].degree), abs(a[i].minute), abs(a[i].second));}else{printf("结果为：%d°%d′%d″\n",a[i].degree, a[i].minute, a[i].second);}degree = (float)a[i].degree + (float)a[i].minute / 60 + (float)a[i].second / 3600;printf("换算为：%f°\n", degree);getchar();getchar();}while(1);
}
void add(ANGLE a[], int i)
{long s1, s2, s3 = 0;s1 = a[i-1].degree*60*60 + a[i-1].minute*60 + a[i-1].second;s2 = a[i].degree*60*60 + a[i].minute*60 + a[i].second;s3 = s1 + s2;a[i].degree = s3 / 3600;a[i].minute = (s3 - (a[i].degree) * 3600) / 60;a[i].second = s3 % 60;
}
void decrease(ANGLE a[], int i)
{long s1, s2, s3 = 0;s1 = a[i-1].degree*60*60 + a[i-1].minute*60 + a[i-1].second;s2 = a[i].degree*60*60 + a[i].minute*60 + a[i].second;s3 = s1 - s2;a[i].degree = s3 / 3600;a[i].minute = (s3 - (a[i].degree) * 3600) / 60;a[i].second = s3 % 60;
}
void multiply(ANGLE a[], int i, int j)
{long s1, s3 = 0;s1 = a[i].degree*60*60 + a[i].minute*60 + a[i].second;s3 = s1 * j;a[i].degree = s3 / 3600;a[i].minute = (s3 - (a[i].degree) * 3600) / 60;a[i].second = s3 % 60;
}
void divide(ANGLE a[], int i, int j)
{long s1, s3 = 0;s1 = a[i].degree*60*60 + a[i].minute*60 + a[i].second;s3 = s1 / j;a[i].degree = s3 / 3600;a[i].minute = (s3 - (a[i].degree) * 3600) / 60;a[i].second = s3 % 60;
}

code blocks中运行示例

如果单次计算的角度多于25个，只需更改这两行代码，改变数组的大小。

    char operation[25];ANGLE a[25] = {0};